A TEXT BOOK
OF
HEAT
C. L. ARORA, M.Sc.
Principal
D. A.y. College, Amritsar
Dhanpat Rai & Sons
JUU,tmr>UR— DELHI
Rs. 12'5i
Useful Publication - -
Mechanics & Properties of Matter— BJ. Kohli
PREFACE
In the preparation of this book meant for engineering and
science students, not only standard syllabus of Engineering degree
classes and A.M.LE. courses have been followed but also syllabi
proposed by other Indian Universities have been consulted. The
subject-matter both as regards the arrangement of chapter as well as
contents of each chapter has been so set that the students may
foiivw the courses easily. Keeping in view this background, the
matter has been explained wherever required with diagrams. At the
end of each chapter expected questions taken mostly from the
various universities, A.M.T.E. and Engineering Colleges have been
provided to make students know the type of questions set in the
examination. The subject-matter has been dealt in M.K.S. systems
and with the latest engineering developments in it. Sufficient
numerical questions wherever necessary have been solved to explain
applications of various formulae.
I am thankful to the Publishers, M/s. Dhanpat Rai & Sons, for
undertaking the publication of this book.
Author
CONTENTS
Chapter
1. Expansion of Solids
Expansion of solids. Linear expansion. Practical
applications. Influence of temperatures on the time
period of clocks. Differential expansion. Thermo-
stat Toluene liquid thermostat. Vapour operated
thermostat. Laboratory method of determining the
co-efficient of linear expansion. Comparative method.
Optical lever method. Superficial expansion. Cubical
expansion. Change of density with temperature.
Effect of temperature on the physical properties of
solids. Expansion of an isotropic solid. Thermal
stresses. Expected questions.
2. Expansion of Liquids
Expansion of Liquids. Apparent expansion. Real
expansion. Relation between Co-efficient of real
and apparent expansions. Effect of temperature on
the density of a liquid. Determination of coefficient
of Apparent expansion (i) the Volume thermometer
of Dilatometer method, (ii) The weight thermometer
method, (ili) The hydrostatic or sinker method.
Principle of U-tube for comparing densities of liquids
Measurement of real or absolute expansion of
liquids. Dulong and Petit’s method. Regnault’s
method. Anomalous expansion of water. Expected
Questions. ^
Expansion of Gases
Expansion of gases. Introduction. Boyle’s Law Lim
tatioDs of Boyle’s Law, Charles’ Law, Gay Lassac’
Law. Absolute temperature. Alternative form o
Charles Law* Alternative form of pressure law c
Gay Lussac s Law. Relation between Y and
Ideal gas equation. General Gas equation. Universr
gas constant. Avogadro’s Number (N) Bolt 2 man
constant {K). Variation of density of a gas To?
done by a gas during expansion. Expected Questions
Specific Heat of Solids and Liquids
British thermal unit (B Th ®Ioric (K.cal
heat unit (C.H.U j. centigrad
Water equivalent Quantity of heat
Pages
1—30
31—49
50-72
(V/)
Chapter
of measuriDg specific heat of solids. Method of
mixtures. BunseE’s Ice. Caloriometer, Joly’s steam
calorimeter, Nernst and Lindermann’s Vacuum calori-
meter method. Method of Mixtures. Joule’s Electrical
Method. Callender and Barne’s continuous flow
method. Method of cooling. Newton’s law of cooling.
Specific heat of liquid by the method of cooling.
E.xpected questions. 73 — JOI
5. Specific Heat of Gases
Specific heat of gases. Two specific heats. Specific
‘ heat ofa gas at constant volume. Specific heat of a
, gas at constant pressure. Relation between two speci-
1 fie heats. Ratio of two specific heats. Change in inter-
1 Inal energy of a gas. Joly’s differential steam calori-
\ I meter for the determination of specific heat at cons-
Wtant Volume. Determination of specific heat at
constant pressure. Regnault's experiment. Scheel and
^ Huese continuous flow method for the determination
of C,,. Expected questions. 102 — 118
6. The Mechanical Equivalent of Heat
The Mechanical equivalent of heat. Thermodyna-
mics. First law of Thermodynamics. Determination
of Mechanical equivalent of Heat. Joule’s experiment
for finding J. Rowland's experiment. Callender and
Barne’s continuous flow method. Callender’s brake
band method. Electrical Method. Searle’s friction
Cone method. Expected Questions. 119 — 132
7. Expansion and Compression of Gases
Expansion and compression of gases. General gas
energy. Temperature equation. Isothermal expansion
of a gas. Adiabatic expansion of a gas. Derivation of
adiabatic equation of a gas. Adiabatic curves are
steeper than Isothermal curves. Work done by a gas
during adiabatic expansion. Isothermal conditions.
Adiabatic conditions. Clement and Desorme’s method
for the determination of y Expected questions. 133—155
8. Kinetic Theory of Gases
Introduction. Kinetic theory of gases. Mean
Velocity or average Velocity. Root mean square
(R M.S.) Velocity. Most Probable Velocity is
defined. Relation between mean Velocity and root.
Mean square velocity. Pressure of a gas. Relation
between Pressure and Kinetic energy. Kinetic
interpretation of temperature. Deduction of simple
Pages
156 >-181
(v//)
Chapter
ypothesis. Degree of freejorn .
iT of enera v^^^ Ratio of specific
Monoatomic gas molecules. Diatomic gas
tnolecuIes^TriatoPiic gas molecules. E^lanation of
ttirce states of” mjttef _6n . the basis, of Kinetic
theory . S'blids, liquids, gases. Mpan fcgg^patli. Calcu-
latiqp qf'^hc mea ^ free path. V iscosit y of__a gas on
T^etic theory. Expected questions.
9. Change of State
Change of state. Latent Heat Laws of fusion. Deter-
mination of melting point of a solid capillary tube or
opacity method. C ooling curve metho d. Super cool-
jp g or super fu siom Effect oFpressure on melting
point Regelation. Freezing point of so lutation.
Mech anis m of^reezin g mix tures, alloys. "Boiling.
Boiling with "bumping. Fffect oriur face tension in the
'forarattoTTDf'bubbles. * Xaws 6T EbuTIit jon or b oiling .
Determination of boiling point. Boiling point of
soliltionsn’nrect of pressure on boiling point. The
triple point. Numerical values for water at the
trfple point. Vapour p ressure over curved^surfaces.
Boiling Formation of clouds. Vapoliniation Factors
favouring evaporation. The nature of the liquid. The
area of the exposed surface. The temperature of
the liquid and air. The renewal of air in contact
with the liquid surface. The pressure on the surface
of the liquid. The dryness of air vapour and gas
vapour. Pressure properties of saturated vapour.
Behaviour of saturated and unsaturated vapours
with the change of volume temperature. Dalton law
of partial pressures. Kinetic theory of saturation.
Moisture in the air. Humidity of air. Absolute humi-
dity. Relative humidity. Mist, fog and clouds. Hygro-
meters. Chemical hygrometer. Dew point hygro-
meters. Wet and dry bulb hygrometer. Hair
hygr^eter. Importance of hygrometry. Expected
qupstfons. 182—223
10. y/Equfttion of State
Andrew’s experiment. General ^
form of isothermals for COg obtained by Andrew’s'^
Y^r Waals jeciuatiqn of a reaLgas^ Vander Waals
equation a nd critic al con stants. Defects in Vander
^aais cquattoi i . ^Rea uce(r-edD^tion of state limn-
T h offison — Pu ^ess. ^Adia batic ex pa nsion Pmr acc
The Pictet or Cascade nroces*;
Liquefaction olf a ir (L i
Refrigerating machines. Domestic ^efrigimor
Expected questions.
Pages
(v//V)
Chapter
11. Thermal Conductivity
Thermal conductivity. Conduction. Thermal conduc-
tivity. Effect of temperature. Temperature gradient.
Ohm's Law for heat current. Thermometric conduc-
tivity or diffusivity. Formation of ice on the surface
of a pond. Conduction of heat through several
bodies in series. Heat conduction through the walls
of a thick cylindrical pipe. Experimental determina-
tion of conductivity. Searle's Method. Lee's Method.
Cylindrical shell method. Lee's disc method. Lee’s
method for good conductors. Cylindrical shell
method. Lees method for bad conductors. Convec-
tion. Radiation. . Black body. Provost’s Theory of
exchanges. Absorbing and reflecting powers. Laws
of black body radiations. Wien's Law.'^Kirchoff's'^
Law. Stefan's Law. Newton’s Law deduced from
Stefan’s Law. Expected questions. 242—282
12. Thermometry
Thermometry. General scale of temperature. Centi-
grade or Celsius scale. Fahrenheit scale (®F). Reau-
mur scale (OR). Kelvin scale (*K) Liquid. Thermo-
meters. Mercury in glass thermometer. Advantages
of mercury as a thermonietric substance. Alcohol
thermometers. Advantages. Disadvantages. Special
thermometer. Clinical thermometer. Six's maximum
and minimum thermometer. Beckmann's thermo-
meter. Standard constant volume. Hydrogen thermo-
meter. Platinum resistance thermometer. Sccbcck
effect thermo-electric thermometer. Selection of suit-
able metals. Base metal couples. Noble metal coup-
les. Fery’s total radiation pyrometer. International
scale of temperature. Expected Questions. 283 — 310
13. Thermodynamics
Thermodynamics. First law of thermodynamics.
Second law of thermodynamic^ R^ey^lbl^ p rocess
ses and irreversible processes. IrreversiFIe process.
Meaning of cycle. The heat engine. Carnot’s cycle
and Carnot’s engine. EflBciency of Carnot’s Cycle.
Internal combustion engines. Spark ignition engines.
Compression ignition engines. Carnot’s theorem.'
Clausius Clapeyson’s equation^ Effect of pressure in
the boiling point. Effect of pressure on the melting
ooint. Otto cycle first stage, second stage, tliird
stage fourth stage. Ideal efficiency. Actual P-V
Diaeram for Otto cycle Suction or charging
^roke, compression stroke, working on power
stroke Exhaust stroke. Theoretical Diesel cycle. An
lt^,ndard. Efficiency of Diesel cycle stearn engine
Rankine cycle. Efficiency of Rankine cycle. Expected
Questions.
311— 336
CHAPTER I
EXPANSION OF SOLIDS
1.1. Expansion of solids. Solids, in general, expand on licating
and the amount of expansion is proportional to the rise of temperature.
The solid expands not only in its length but its surface area and
volume also increase. The expansions of length, area and volume
are respectively called Linear expansion, superficial expansion and
cubical expansion.
1.2. Linear expansion- When a rod of a solid material is heated, it
IS found that its length increases, the increase in length is called Linear
expansion.
j length of a rod at 0°C. If the rod is Iieated to
Its length becomes L|, then '
Increase in length^ I'o
It is experimently found that the increase in length is
{i) directly proportional to the original length Lq
{ii) directly proportional to the rise of temperature t and
{in) depends upon the nature of the material of the rod.
Combining all these factors we have
or
•w
—1^0 (l
Where a is a constant depending upon the nature nf fU
and IS called the coefficient of linear expansion ^ material
From (i) we have
a
’o_
Increase in length
original length x rise'in te5ip:-= ^
1
• • »
(u)
2
ENGINEERING PHYSICS
If Lq = unit length and then
ot Increase in length per unit length at for 1®C
1 X 1
rise in temperature.
The coefficient of linear expansion of a solid is a veiy small quantity
so it is not necessary that initial temperature may be 0®C. Hence the
coefficient of linear expansion of a solid may in general be defined as
the increase in length per unit length per degree centigrade rise in
temperature.
If the length of solid is kno^vn at an initial temperature (ptber
than OV) its length at any other temperature f/C can be found o
as follows.
Let Lj = length of the rod at /jT'
= length of the rod at t^C
From equation {U) we have
Lj = I-o (• "f
and similarly
L2 = -^0 +‘^^2)
L.y _ 1
I
= ( H-a/2) (1
= l+a /2 — a/,+ terms containing a- etc. which are very
small and can be neglected.
= H-a(f2- ^ 1 )
... =
and
...{Hi)
...{iv)
From equation (/r) it is clear that thcj-fficient^oflinear
efficient of linear expansion per centigrade degree.
EXPANSION OF SOLIDS
3
Table of coefficients of linear expansion
Substance
a pct®C
Aluminium
24x10-®
Copper
16 5x10-®
Brass
18 9x10-®
Cast iron
IMxlO-®
Stainless steel
10-4xl0“«
Mild steel
121x10“®
Invar (alloy of nickel and steel
containing 36% Ni)
0 9x10“®
Nick el steel
13 0x10-®
Silver
13-8x10-®
Gold
140x10“®
Tin
21-4x10“®
Lead
27-6 X I0-®
Platinum
8 9x10“®
Tungsten
4xl0-«
Glass
9-5 X 10“®
Gjass pyrex
30xl0“S
Duralumin
22-6x10-®
an iron ,‘od are each one metre long at
VC. Find the difference in their lengths at 100°C One>er
^imnsxim of brass is 0 000019 and that, of ironLo 000012^ ^
Length of each rod at 0®C
= 100 cms.
Rise of temp.
/=I00''C
Length of brass rod at 100®C
Length of iron rod at lOO'^C
=Lo(i+a/)
= 100 (1+‘000019 X 100)
= 100'19 cms.
= 100 (1+0*000012X100)
= 100*12 cms.
i
4 ENGINEERING PHYSICS
Difference between the lengths of the two rods at 100®(7
= 100 19— 100 12
=0 07^cm.
Example 2. ^ disc at 'JO^'C has a diameter of 30 cm. and a
hole cut in the centre is 10 cm. in diameter. Calculate the diameter of
the hole when the temperature of the disc is raised to oO'^C. The coeff^
of linear e.i'2}ansion for hrass — O'OOOOlS.
Let the circumference of the hole at 20®C be
y 10
Lj = 2::/-=27:X — = 10- cms.
Suppose after heating, the rad'us of the hole becomes i?, so*
that the circumference of the hole at 50®C is Lt=‘2-r:R-
Now [!+;>'•
2-i?=107: [1 +0-000018 X (50— 20)]
or 2i?= 10 00540 cm.
Hence diameter of the hole at 50 ®C= 2^=10*00540 cms.
Example 3. A metal rod of diameter one cm. measures 50 ^s. ^n
length at 20°C. it is heated to 05^C. its length becomes oO’OO cms^
What is the coeff. of linear e.rpansion of the rod. What u'ill be the lengt
of the rod if it is cooled to 0°C.
Length of the rod at 20'’C = 50-0 cms.
Length of the rod at 95°C =50*06 cms.
Increase in length =50-06 — 50-0
= 0-06 cm.
Rise in temperature =95 — 20 = 75°C
increase in length
Coeff. of linear e.xpansion «=onginal lengthxrise in temp-
= . =0 00001 ^
sox 75
Let Lq be the length of the rod at 0°C, then
//i=-Z>o(l +^0
50*0
1 +0*0000t6X 20
=49-987 cms.
EXPANSION OF SOLIDS
5
13. Practical Applications. The expansion of solids is
utilised in several ways, in some cases it is a welcome effect while in
others it is an adverse effect to bo overcome. Some of the important
applications where it serves a useful purpose art' : —
(0 Sealing wire in glass. The expansion of ])latinum is very
nearly equal to that of glass. Hence wires of jdatiinim are used for
sealing into glass. If we use any other material its expansion will
be different from tliat of glass. Hence when glass cools after
sealing of the wire, it will crack due to unequal exjiansion. Xow
a days special alloys of steel and nickle (55'!o + 45^o) liave lieon
made of wliich the expansion is tJie same a.s that of glass. These
are comparatively cheaper and have rejdaced platinum for tliis
purpose.
(ii) Prevention of bulging walls from collapsing. The expansion
and contraction of iron rods is utilised to prevent the bulging walls
from collapsing. A few iron rods are passed througli outwardly bulging
walls across che building and are lieated at the centre. The
extremeties of the rods are then tightly fixed by nuts and bolts to
iron plates secrewed to the wall, on the outside. Wlicn the rods cool
and contract, tlie walls are pulled together and assume their proner
positions. ^
..11. Riveting of boiler plates- The boiler plates arc rivetted with
red hot ruetS; These on coolmg contract and grip the plates tightly.
The joint IS thus steam tight. ‘•*5
o Tu To fix an iron tyre on
a wooden wheel, the tyre is always made of a diameter a little Jess
than that of the wheel. The tyre is heated and. as a result t
expands. It is then made to slide over the wheel. On coo if
contracts and grips the wheel ver^^ tightly, ^
+ ■ 1 + Sometimes glass stoppers stick in the neck of the bottle too
neck ^of ordinary means. To remove the stopper the
neck of the bottle is carefully heated. The neck evnandQ k,,.
heat is conducted t^it^ As^a
result the stopper becomes lose and can be easily opened. ‘ ^
(ri) The differential expansion of two solid<i i- • *i
construction of bimetallic strips us#*H in .k ^
These have been discussed in Art! thermostats and relays,
f.l i“"l 5 i. . h.™-
^le rails would expand and bend thereby c^sine^!l
bolted to them is sho4 in the Fig t . Th.
the bolts joining the rails pass are Lai in sLpe Twraflowf the
6
ENGINEERING PHYSICS
movement of the two rails in the direction of their lengths due to
a change in temperature.
Fig. 1.1.
Hi) Similarlv while constructing a railway ^ndge Provis^^^^^
must be made to accomodate the variatioiyn \
to summer to prevent any damage to the bridge and its pieces,
achieve this the^ two ends of a long bridge are set on rollers so that
the change in length may take place without damaging the
{Hi) Pipes carrying steam are given loops along their leng
allow for change in length.
(;.) The telephone and electric ^ures between the po|es^ahra>^
ret la.T^s’’t;^°Xw‘r ’Jht etant; and contrac tlon.
. .. <‘V£' 'sf-A^'Assv-orr'S -i srs'sl
fSatrsMily
reading becomes high.
If a scale measures a length I-j
perature //C the length measured as would have a 6
X,2~ ^i[I +*(^3 — h)]
true length=Observed length [H-a{rise of temp.)] ^
And correction to be applied to the observed length=obser\e
length xaX rise in temperature
^pansion of which with rise in temperature .s very
Example 4. A tape vsed f^r measuring
and is correct at a temperature of lo C. If the tape us j
EXPANSION OF SOLIDS
a distance, of 700 metres when the temperature is J0°C. what will he the
total error in the measured distance due to expansion of the steel tape \ [a
for sfeel=13 X 10~^ per^C.]
We know that
L 2 =Z<i[l -f a{/2 — ?i)]
Where 1/2= Actual length at 15°C and
Li=ohseTved length at 10 “C
700 ~Li [14-13 X 10'® (10 — 15)]
=observed length (1 — 65X10“®)
or observed length^
700
-6\-l
= 700(1— 65 X I0-«)
(1— 65X 10-«)
= 700 (1+65X 10“®) =700400455
= 700 0455
Error=+0 0455 metre
(ft) Concrete roads and floors are never laid in one piece.
These are laid in sections and expansion channels are provided between
the sections. Without these expansion channels, the concrete would
soon be broken up due to the large forces caused by the expansion.
(uu) A thick glass tumbler usually cracks if very hot liquid is
poured into it. The reason is that inside of the tumbler gets heated
and expands even before the outside layer just becomes warm. Thus
unequal expansion of the glass produces strain in the glass and this
causes the tumbler to break. In a thin-walled beaker, however there is
uniform heating and uniform expansion everywhere.
The expansion of pyrex glass is about ^rd that of expansion of
ordinary glass. It is for this reason that the pyrex glass can be
heated without cracxing.
<=
V
(«m) The moving parts of machinery such as axles and shafts, are
gaurded against their expansion due to the heat produced by the use
of lubricating oils.
(ta:) A pendulum is used in a clock to keep time. We know that
the time period (time of oscillation) of a simple pendulum is given by
— where I is the length of the pendulum and
acceleration due to gravity. The pendulum used in a clock fe'i»l>t a
sunple pendulum but is a compound pendulum consisting of a metal
bob^ suspended by a thm metal rod. Hence if the length of the
penaulum is increased due to rise m temperature, the time of oscilla-
tion increases and clock loses. If the length decreases due to fall in
temperature, the pendu urn swings faster, the time of oscillation
becomes less imd the clock gains. Thus a clock wUl lose time in
“s^^r'electSe^Sansior**^ - compensated
There are several methods of
compensating the pendulum for
s
ENGINEERING PHYSICS
IRON
chang:e in temperature. Probably the best known is Harrison's
grid- iron pendulum- The principle of this pendulum is explained below.
Let two parallel rods AB and CD
mad e of two different materials,
iron and brass of different lengths be
joined by a cross bar BC. The system
is suspended at A and the end D
carries a bob which is free to vibrate.
With the rise in temperature, the rod
AB will expand in the downward
direction while the rod CD will expand
in the upward direction. If the lengths
of tlie rods are such that the down-
ward expansion of AB is equal to the
upward expansion of CD. the centre
of gravity of the bob will remain in
the same position and, therefore, the
effective length AD of the pendulum
will remain unchanged.
Fig. 1.2.
If L is the length id iron rod *4^ and that of the brass rod CD and
a, and a, tlieir coefficients of linear expansion respectively then in order
that their expansion may be tlie same for any rise of temperature.
f 1 j / 2 31 2^
or
^ 1^1 — ^ 2*2
/j X.>
/
Hence, for compensation, the lengths of the two rods should be
in the inverse ratio of their coefficients
of linear expansion.
Harrison’s grid iron pendulum is
constructed on this principle. It
consists of a number of brass rods
(shown by double lines) and steel
rods (shown by thick lines) arranged
alternately on a common frame suspen-
ded from the point A. There are 9 para-
llel rods out of which 4 are made
of brass and 5 of steel. The bob is
carried by the central steel rod. The
ends of the rods are attached to cross
bars in such a way that the steel rods
expand downwards and tend to lower
the position of the bob B while the
brass rods expand m the up^'a^d
direction and tend to raise the bob B. In
order that the pendulum may keep correct
brass
5TC£L
Fig. 1.3.
time the distance bet-
EXPANSION OF SOLIDS
9
ween the point of suspension and centre of gravity of t]ic ])ciulu)um
should remain the same at all temperatures.
As is clear froir Fig. 1.3 all the rods ox«e])t the central
rod are in pairs. If there are 5 steel rods and 4 hr.a-^s rods, tiu*
effect of five steel rods is tl»e same as that of tlin-c and the
effect of four brass rods is the same as that f)f two. Tlie ]n,-ndulum is
I 9
so designed that the length of the three steel rods is - times the
length of the two brass rods. Since the coefficient of e-xpan^-ion of
1 9
brass (0-000019) is -y^times that of steel (0-00001 2),. tlieir e.xpansions
are equal and opposite and the centre of gravity of the bob remains
in the same position.
A compensated pendulum used in clocks to kee]^ correct time
cannot be used in a watch or a time ju'ecc. In these an arrangement
known as Balance Wheel is used. It essentialh- consists of two or
three curved segments as shown in the Fig. 1 . 4 . The wheel is sujqiorted
i-ig. 1.4
on a vertical axis and oscillates under the influence of the hair
spring S. The tune period of the balance wheel is given by
V
c
[i] Where I is the moment of inertia of the balance wheel which
111 turn depends upon the radius of the wheel
10
ENGINEERING PHYSICS
higher coefficient of linear expansion like Brass is used on the outside
and iron which has smaller coefficient of linear expansion is used on the
inner side. The rim is made in two or three segments, each part being
supported by a spoke of the wheel. Each segment of the wheel also
carries small sccrew weights.
When the temperature rises, the increase in the length of the
spoke causes the points ,4 to recede from the centre of the wheel
and the elasticity of tlie hair spring also becomes less. As a
result the balance wheel tends to move slower. But at the
same time, the free ends of the segments of the wheel curve inward,
thereby decreasing the effective radius of the wheel and hence
wheel tends to move faster. These two tendencies, one making the wheel
to move slower and the other making it to move faster, are made
to compensate each other so that the time period remains the same
at all temperatures. Exact compensation is obtained by adjusting
carefully the screw weights on the three segments of the wheel
the screws near the free ends tend to increase the compensa ion
while those near the fixed ends produce an opposite effect. Uniy
precision watches such as chronometers and wristlets are proviaea
with compensated balance wheels.
With the introduction of Invar steel in the manufacture of
pendulums and balance wheels the problem of
been eliminated to a great extent, on account of
small coefficient of expansion of this alloy. Since however, invar
has a rather high magnetic susceptibility and is / 1^1
in earth’s magnetic fields, other alloys such as J • t
nikel alloy) of smaller suceptibility and of f..„*ches
of expansion havebeen produced and utilized. In modern
balance wkel is made of inoar having a hair spring made of ehnvar
{steel-nickle alloy).
1.4. Influence of temperature on the time period of clocks.
The time of oscillation of a clock pendulum is given by
■Where i=length of the pendulum.
j=acceleration due to grav'ity.
When temperature changes, the length of the pendulum of a clock
changes and hence the time period is also affected.
At temperature the time period
J
...(»)
EXPANSION OF SOLIDS
At temperature < 2 * time period
Dividing («) by (i) we have
^ j "ir ...(itt).
Ti V h
Now ^2 = ^1 [1+“
=1+C^ (*2 M
*1
Substituting in equation (m) we have,
^ =[i
a(i 2 — <i)] neglecting squares and higher powers-
of a(i 2 — /i)
or «(<*— <i)
or a(f2— <0
or ^=ia •••(•■«’)'
^=Fractional change of period with the change in temperature..
Hence loss or gain per day—dT X 3600 x 24 seconds.
If the value dTjT is positive the clock will lose time and if it is
negative it will gain time.
Example 5. The pendulum of a clock is made of brass whose
coefficient of linear expansion is TO X 10~^ per degree. If the clock keeps-
correct time at 20°C, how many seccmds per day will it lose at Prove
any formula used. {A.M.l.E.)
Fractional change in time period when temperature changes from-
20®C to 35®C.
=4
=JX 1*9 X 10^ (3S— 20)
=iX 1*9 X 10-5X15
= 14*25X10-5
dT
T
12
ENGINEERING PHYSICS
This means that the clock loses 14-25 X seconds for each
second interval. Hence the loss per one day would be
= 14-25 X 10-*X 3600X24.
= I2‘3 Seconds
Example jS- dock which keeps correct time at ‘20°C has a
wmhdnm made of brass. How many seconds will it gain day
when the temperature falls to freezing point ? a for brass— O-OOOOW
per °C.
Fractional change in time period when temperature changes from
20T to O^C.
1 if
Y (^2—
As is negative the clock will gain time :
‘ Gain in time per second.
— i a {tn ^l)
= ^X 1-9 X 10-5 (0—20)
— _19X 10-5
Gain in time per day =3600 X24X(— 19X10 5)
= — 16*416 sec.
Example 7. Find the change in the period of
.mnduXum of thin brass wire when the temperature changes from 20 Cto
iflhe vendulum is 122 0 cm. long at the former temperature.
OS l^rass is 10^10^^ pcr ^C. Given that
.g^OSO cms.lcm^.
Fractional change in time period is given by
— a [ti /ii
=4X1-9X10-^ (25-20)=4-75 XI0-5
Now T
j
27r
122
980
= — = X 1*12 = 2*23 sec.
\/l0
• Change in time period
dT—4-15 X 10-5X 2-23 =10-6 X lO"® sec-
Example 8. A 71 iron dock pendulum Mm
-one day. ^At the end of the next day ^ for
the change in the temperature. Coejf. oj L,me r
'iron^l-irxlO-” per°C.
13
expansion of solids
The number oi seconds in a day=86400 seconds.
On the first day the clock pendulum makes 86407 oscillations
86400
Period of one oscillation 1
On the next day the clock pendulum makes 86407 12=^86395
oscillations
86400
Period of one oscillation A—
Now for a pendulum. 7^
where ?i=length of the pendulum on the first day at liC and
a/
= V = [1+*V'2
A)1 ^
86407
86395
86407
86395
= 1^1+ “ t^2 M J
= l+4x 1-17X 10-5X
86407
— 1
^ # 86395 12X 2X 10*
... Change m temperature ^ ^ 8,395 X 1 • 1 7
= 23-8‘^C
Example 9. A brass penduluni hteps correct time at 12‘^C. If it
loses 6 seconds in a day when temperature is 19 (>*^C , calculate the
coefficient of linear expansio?i.
dT
The fractional change in time period-^ — ^ 1 )
==|a(l9*5— !2)--ia7-5
Now Loss per day =dI'X86400 = 6
dT=
® 86400
As the clock keeps coirect time at 12®C
T—l sec.
^ dT ,
and -^=Ja7-5
•••(f)
86400
= |a7*5
Hence
a =
86400X7.5 I8'5XI0®/C
14
ENGINEERING PHYSICS
1.5. Differential expansion. From the table of coefficients of linear
expansion, we find that the coefficient of linear expansion of brass is
necirly % times that of iron so that
a brass strip will expand % times as
much as iron when heated through
the same range of temperature.
Consider a strip of brass
and iron rivetted together
so as to form a composite bar.
If this bar is heated, it will be
found to bend because of the un-
equal e.xpansion of the metals.
Expansion of brass being greater
than that of iron, brass will lie on
the outer side of curve when hot.
Similarly when cooled the brass
will lie on the inner side, its con-
traction being greater than that
composite bar is known as a bimetallic strip.
Bimetallic strips are used in metal thermometers, in balance
wheels used in watches and in thermostats for regulating the tempe-
rature and also for controlling many electric devices by temperature
cli3^n*^cs
16. Thermostat. A thermostat is a device containing a bime-
COLD
Fis. 1-5
of iron. Such a compound or
M ntr-f5
HEATING COIL
'^/VWVVWV
RESISTANCE WIRE
tallic strip which
Fig. 1.6
is used for maintaining any temperature constant
£XPANSION OF SOLIDS
15
at a given value. The expansion or contraction of the bimetallic strip
can be made to make or break two electrical contacts at the desired
temperature. The thermostat can be used to control the temperature
in a referigerator, or an eletric iron or a room heater. Let us see how
it helps to control the temperature in an electric iron.
The thermostat consists of a bimetallic strip of brass and iron fixed
at the end A and free to move at the end .B which is in contact with
the screw 5 . When the arrangement is connected to the supjily
mains through the heating coil H, the current flows, .-^s soon as the
heat developed in the heating coil reaches the desired maximum
value, the bimetallic strip also gets heated and curves upward due
to the unequal expansion of bass and iron. This breaks tlie circuit
and stops the flow of current. On cooling, the bar straightens again, and
the contact is restored which completes the circuit. The temperature
is thus maintained constant.
The use of a bimetallic strip in automatic Fire Alarms is
shown in the Fig, 1.7. The action of the fire alarm is
based upon the bending of a bimettalic strip on heating. Such type
of thermostats are placed at many points in a big building where
there is a danger of fire. The end B is connected through a bell and
a battery to the screw S which is not in contact with the end A.
If fire breaks out or temperature at a given point in a building uses
beyond a certain point, the bimetallic strip gets heated and bends.
This makes the contact of the screw S and thereby tlie bell circuit is
completed. An alarm is thus raised by ringing of the bell.
Fig. 1.7.
fnii K ^ Liquid Thermostat. If a glass bulb containir
toluene, benzene, alcohol, or some such substance possessing a Ian
coefficient of expansion, is placed in a bath of wateror oil it
r\ihirh^^- % ***®™®?t*L Toluene thermostat consists of a loAg tul
tntlct ^ The liquid^ is
contact with a mercury tube M. The bulb T is immersed in tl
16
ENGINEERING PHYSICS
bath whoso temperature
is to be maintained constant. The bath is
heated by a burner and the gas supply
for it enters through the tube x, flows
through the tube y as shown by
arrows. As tlie temperature rises, tlie
expansion of the toluene in the bulb
T forces the mercury up the tube M
until it closes the opening of y and
stops the supply of gas reaching the
heater. Actually a small side tube Z
allows some little gas to pass in order
to keep tlie flame burning, but the heat
supplied is not sufficient to maintain
the temperature of T. Hence, the
toluene contracts, the opening of y
is increased and more gas passes to the
burner to heat the batli. By regulating
the amount of mercury in the reservoir
R the temperature of the bath can be
kept constont at any desired value,
accurate to one or two degrees.
Iir the case of electric heater, the
tubes at .r and y are removed, and
- their places are taken by two platinum
.vires, O.;;?' .vlro -aled -y
the mercury surfaces. hath is kex:)t well stired.
T+ imnnrtance to ensure that the oain is>
-"-..cted
bath can be kept con.stant accurate 0 0 i t . biicn a
be used up to a temperature of about 100 C. ^ ^
1.7. (h) Vapour operated thermostat. The principle
TO
MOTOR
Fig. 1-9
such a thermostat is often used -in referigerators as shown in Fig
1.9
EXPANSION OF SOLIDS
17
In this fomi of thermostat a volatile liquid is filled in a bulb which is
placed in the cold chamber of the refrigerator whose temperature is
required to be controlled. The change of temperature causes the cliange
of vapour pressure in the bulb which in turn causes expansion and
contraction of matallic bellows {MB) which further makes and l^reaks
the contacts of the electric motor of refrigerator. Kow, when the
temperature of the chamber rises, the increa'^e in vapour pre^siire
is communicated as shown in the capillary tube into the metallie
bellows, makes the electric contacts and thus starts the refrigerator
motor.
When the temperature falls below the required value, the
metnllic bellows contract due to decrease in vapour pressure which
breaks the contacts and stops the motor.
1-8. Laboratory method of determining the co-eflicient of linear
expansion. A simple method commonly used in the laboratory to
determine the co-efficient of linear expansion mukes use of the ajuia-
ratus shown in fig. no.
The apparatus consists of a lod CD of a given material
about 60 cms. long enclosed in a wide tube supported on
two supports ft.xed on the base of the apparatus, Tlie ends of the
outer tube are closed by means of corks tliroijgh which pass the ends
of the rod CD. One end of the rod is m ide pointed and tite
other IS kept flat.
eJg. 110
The outer jacket is provided with tivr. n j
serving as inlet and outlet for steam so t ^
required temperature. Through the ceiUraT ^
T IS inserted to measure the temDeratiirf ^ t^ermomet
rod rests against a rigid suppT Xe the ot?'*' ^ t'
tip of the screw of a spherometer suonorted ^ touches tl
in such a manner that the screw and rod areln^' thl^Tame^llr^lgi
IS
ENGINEERING PHYSICS
lino. Tlio rod i>. thus free to expand only towards C. To detect the
oxjct contact of tlio secrow witli the end C of the rod, a galvano-
inetor (/. and a cell C are connected through key K as shown.
Tlic lengtli of the rod CD is measured as accurately as possible
with the help of a metre scale at room temperature. It is then
]>laced in position in the jacket tube and the temperature indicated
bv the theromoiueter is noted. The key K is inserted and the
‘'j>horometer screw is moved gradually till the galvanometer gives a
sudden detlection. The reading of the splierometer is then noted. Now
tlu‘ screw is turm'd back and the steam is allowed to enter the inlet
tube I. .\fter j)laying round tlie rod (7>,the steam escapes through the
outlet tube O. The temperature of the rod CD goes on rising, as indica*
t< d by tlu rmometer T and finally becomes constant. After waiting for
some time, to ensure that the rod has attained a constant temper-
tlirougluut. tlic splierometer screw is again moved
atur
to touch the end C .md the reading is noted in a similar
manner. Tlie final reading is rej'eated again and again till it is
constant. The dilference between these two readings of the sphero-
meter gives tlie Increase in length of the rod for the rise of temperature
during the ex]H‘riment. The co-eofficient of linear expansion is calcu-
lated by using the formula
original
change in length
length xrise in temp.
ix (100— n
^i^ = changc m length of the rod.
i> = original length of the rod at room temperature CC
and (100 — /) is the rise in temperature.
1.9. Comparator Method- Ihis is a standard, precise and a
direct metlu’d employed for determining the coefficient of tnermai
expansion when the specimen is in the form of a bar, rod, wire or
tube. The principle involved may be stated as follows .
■■Two fine marks are made on the experimental rod. A standard
rod. one metre long and maintained at a constant temperature is
used to measure the distance between these two marks of the
experimental rod which varies as the rod expands. Two microscopes
provided ^vith micr.rneter eye-pieces or micrometer screws are used m
the comparative estimation of expansion. Hence the method is
known as Comparator Method-
The rod under test with two fine marks, A A at a certain distance
apart and a standard metre are mounted horizontally in two double-
walled troughs, arranged parallel to each other. The troughs are
fixed on a platform running on rails, so that either the rod
standard metre can be brought into the field '
scopes mounted vertically in the two massive pillars PF.
To begin with, the troughs are filled with water and melting ice
is placed in the space between the double walls. The experiment
EXPANSION OF SOLIDS
19
rod and standard metre are both allowed to attain the temperature
of melting ice 0®<7. The standard metre is brought underneath the
Mg. 1.11
microscopes. Tliese are adjusted using the micrometer screws until
two lines of graduation on the metre, approximately tlie same
distance apart as the marks on the other rod. are in coincidance with
their vertical cross-wires. Next, the experimental rod is wheeled into
position and the microscopes are set on its marks. From the chance
found i ' «
Now, the ice cold water in the trough is renlaced hv • \
can be heated under thermostatic control «o ’ ■ * "hicii
can be maintained at ' T^h^rnT"*"'
and the two mmroscopes will have to be moved in order to 'focur‘'tf "
cross-vires of their eye pieces on the two marks focus the
Imear expansion can be calculated by appiyin^theTefatior'^'"'"’^
a
Lnt
where AL is the change in length as measured by microscopes
Lq — length of the rod at 0®C
f=rise in temperature.
dispensed wrth!‘and\iaterl't®coi^^an^^ the water baths may be
circulated through the inside of ?he tuhl be
measured directly but it is magnified bt mL ‘ ® material is not
The apparatus is shown in Fig.^I 12 Tl^ 2 d optical lever.
the end The free end I
20
ENGINEERING PHYSICS
top ol which is attached Ihc telescope at right angles to rod BO. The
telescope TT is horizontal and is focussed on a distant vertical scale
S. The bar AB is first immersed in ice cold water (0°C) and the scale:
is observed tlirough the telescope. Let the scale division visible be-
t
Fig. 1.12.
E, The bar is then enclosed in a constant temperature water batlv
which i .5 kept at some known high temperature The bar expands-
and the free end B pushes the rod BO at the end B to say C. This^
causes the telescope to rotate into a position given by T'T' , the new
scale reading F is then observed through the telescope. The actual)
small e.xpansion BC of the rod is obtained from the relation,
EF BO
BC=BO. tan BOC=BO-tan EOF=BOx =^.EF
Since BO, OE and EF can be measured, SC is determined with:
fair accuracy by the magnifying device employed.
The coefficient of linear expansion a of the material of the rod is.
then given by
BC
BO jrrx-L-
OE‘^
LXt U£j J-'O
Where Z.„ is the original length of the rod at 0°C, and t is the
temperature of the hot bath.
The same idea may be applied in a simple way by using instead
r rretc/le. ^h^e-Se‘:?t^fbUf tuZ:'lS^ ^
t3“t\r eUsmn
can be calculated* ,
the InVreasftf s^u^f trea "oTSg. “"the linear
expaShe superficial expansion also depends uporr
[i) the original area
EXPANSION OF SOJDS
21
(ii) the rise in temperature and
(n’t) the nature of the material.
If So be the surface area of a metal
plate at temperature 0®C, and St its surface
area at temperature as shown in Fig i . 1 3
then, as in the case of linear expansion
St~So=^ Sot ...(0
Si—So-\~^Sot~So ...(ti)
where 3 is a constant depending upon the
nature of the material and is known as the
coefficient of superficial expansion.
From (t), we have
R= Fig. 1.13.
Sot
Hence the coefficient of superficial expansion is defined 05 the
increase in surface area per tinit area per degree centigrade rise of
temperature.
If Si is the area of a surface at a temperature and S^ at a
temperature < 2*^1 then
S^=Si [I+P ih-h)].
1 ' 12 . Relation between a and p. Let ABCD he a square metal
plate having each side of length lo at 0®C, Thus its area at O^C.
So^lo
when the metal plate is heated through ec, the length of each
side becomes It as shown the Fig. 1 ‘ 13 .
Surface area at t°C, St==lt^.—[loH-^<x.t)]^
where a is the coefficient of linear expansion.
As already defined, the coefficient of superficial expansion
a= gciga. ^ lU 1 ?„2[(l+a()2_i]
laH T^t
■ B= I + .
t I ^ =a-/+2a.
Since a is small, the value of a* can be be neglected
Hence (3=2a.
tw Jfi; cS,;;**, i, "p*"”"* »> »
the cubical expansion. Like th” linear Lh T-
the cubicial expansion also deperids upon s^PerScial expansions.
(i) the original volume
/-•X temperature and
the nature of the material.
22
ENGINEERING PHYSICS
Let Vo be the volume of a body at 0°C and Vt the volume at some
higher temperature. t°C as shown in the Fig, 1.14 then change in
volume
••■(0
V t — 1 o — ot
or r,=ro(l+y/)
where y is a constant depending upon
the nature of material and is known as the
coefficient of cubical expansion of the material
of the body.
From (<) we have
1 f — i o
— rg- ^
Therefore, the coefficient of cubical
expansion is defined as the increase in Fig. 1.14.
volume j>er %init volume per degree centigrade rise of temperature.
1 - 14 . Relation between a and y. Consider a cube of side 3^^
temperature (VC, then its volume at 0°C.
— 7 3
I 0 — Iq
Let it now be heated to VC so that each of its side becomes h
as shown in Fig. 1*14 and its volume becomes F£=/t®= [/©(I +«£)]*
We have
Vt^Vo _ M i +a0]®—
r y--( Ut
0~l ^O^t
1] _ ( l+a<)3— 1
“ lo^ t
“ t
Since a is small, the terms involving «=> and can be neglected,
SO that
^This shows that the coefficient of cubical expansion is three limes
the coefficient of linear expansion.
1.15. Change of density with temperature. We know that
_ Mass 3/
Density P=v^ii^e“ V
XT nQ thp volume of a substance increases on heating and
,h. S
is raised.
expansion of solids
Let a given substance of mass M be heated from 0*C to i C.
There is no change in the mass M , but its volume changes from Vo at
0°C to Vt at t°C and density from po to Pi. Thus we have from
definition.
M , M
Po= and Pt^jT-
\ o ft
Pt Vo Vo
Po=P( (i+yt) ...(0
(Note. Density decreases with rise of temperature.)
Hence
Po~Pt
Pit ’
The above equation gives us the relation between coefficient of
cubical expansion y and density, when the temperature rise from O^C
to eC. If the temperature rises from to and and P are
the densities at and respectively, then from equation (»)
we have
Po — Pt\ (l+yfi) ...(«)
and Po=<>/s(I+yy ...(m)
P(i (•+y<i)=P/2 (H-y<a)
or P( 1 +P(, yii=Pti+Pi,Yii
or = y
(P(2 '>-P<l«l)
Hence knowing the densities of a substance at two different
temperatures ti and ig, the value of y (coefficient of cubical expansion)
can be determined indirectly. ^ '
many other properties of a solid which"chSe °Wth°temrerature.'^ ""
1 . Density, (p). It has been proved in Art I*IS that p^^p, (i-fy/)
Where p,, and p^ are the densities of the bodv at anH t^rr
respectively and y the coefficient of cubical expansion. ® ^ ^
depends ““sla^'Tnd IS' diSf of a body
its mass U rem“fn"eonsfanf, “'C^urdimens"^
increase, consequently the moment ol i“‘° wm‘ nSefr^i
24
ENGINEERING PHYSICS
familiar example of this fact is afforded by the balance wheel of a
watch which has to be compensated against this increase m its
moment of inertia.
3 Elasticity T^be variation of elasticity with temperature
does not yield readily to theoretical treatment, but experimental
investigations show that in general it decreases with rise in
temperature.
1.17. Expansion of anisotropic solids- In the case
substances the expansion due to heat is usually different in different
directions. Such a body is called noji-isotropic. In every crystal
there can be found three mutually perpendicular ‘^'"^tions such that
if a cube of a crystal is cut with its edges parallel to the directions,
then on heating, the angles will still remain right angles although
the lengths of the sides will change unequally. These three directions
are called principal axes of dilation of the crystal and the linear coeffi-
den^s of expansion along these directions are called the Principal
coefficients of expansion and may be reprsented by a^, ana a^.
Consider a cube having initial volume V, and initial length of
side /ft.
Then
1 ’ — / 3
* 0 — *0
.\fter heating, the new volume I’ wiU be given by
r=[/„ (i+»ii')] ['o + ['o (1+“^')]
= ( 1 + M ) ( 1 + V )
=Fo [l + (*x+«!/+“=)
Thus we see that for an anisotropic crystal, the coefficient of
cubical expansion y will approximately be given by
y=«.r+* 2/+^-2 ^ . .
e The coefficient cf cubical f ^
by the algebraic sum of the three linear coefficients.
Note In some crystals, one of the coefficients may be negative,
^otropic bodies are thc.se whos^ e^'!^
Example to. .4 circular l^UW cm. in IZcToTf'^
copper=16’6x20 ^ per C. • * Viv
The area of a surface S. at a temperature C is given y
[1+P (ta-y]
25
{EXPANSION OF SOLIDS
where 5 , is the area at temperature /i
Increase in area= 52 — 'Sj
10 10
Now ^i=r:x ^ ^
and p=2a = 2Xl6*6XlO-« per
I02x2X16-6x10-«X(125*-75")
= JL 102X2X16-6X10-<'X50
4
=0*13 sq.cm.
4
Example 11. If the volume of a block of metal changes by O' 12%
-when it is heated through 20^C, what is the co-efficient of linear exjpansion
■of the metal I
The volume at i ®C is given by
Vt^Vo(i-hyl)
Vt-Vo
or
Now
lV/«
0* 1 2% = 0*00 1 2
1 0 ‘
Change in temperature t=20
4 4
0*0012
= a 0*00006
But
20
y=3a
y __0*00006
“33
= 0 00002/®C
Example 12 . Density of a substance at O^C is 8'92 and at SO^Cia
^'80. Calculate the coefficient of its linear axpansiwi.
Density at Po=8'92
Density at 80®C, p8o=8*80
Using the relation Po=P< (l+yO we have
or.
Po
Pt
8*92
8*80
= l + y(=l-i-y80
= 14 - 80 /
26
ENGtNEERING PHYSICS
or
or
But
_/8'92
1
8-80 80
y=s0-00017045
y=:3a
y _0*00017045
“3 3
= 0 00005682
Example 13. A piece of a certain metal weighs 50'5 gms. in air
46'0 gms. when immersed in a liguid at lO^C and 46'38 gms. at 45'^C.
Find the coefficient of cubical expansion of the liquid if the coefficient of
linear expansion of the metal is 2 x 10-^ per ®C.
Loss of weight at l0°C=S0-5-46=4-5 gms.
Loss of weight at 40°C=50-5-46*08=4-42 gms.
By Archimedes principle, we have
Loss in weight=Volume of the metal X density of the liquid
4'5=f^io Pjo
and
4-42 = ^40 P 40
#
• •
Pjo 4*5 ^^40
P 40 4*42 K,o
But
or
(40_10) = 14-30 y
But
8
It
•
• •
!^=1 + 30X 3x
^10
^1_|_30X3 (2X I0-=)
»
• •
Substituting in (i). we have
111- X (1+90 X 2 X 10-®)
P 40 4-42
= 1 02X 1*0018
...(»*)
But
9
0 •
Pl0=^40
=P4o (i-hyX30)
?i9=l+30y
P 40
EXPANSION OF SOLIDS
•27
Equating (a’) and (tu), we have
-i-30y= 1-02 X 1*0018
1 02X 1-0018— I _r05 — l_0-05
^ "30 ^ 30 30
= 1-66 X 10-®
Example 14 . A hollow cylindrical vessel is heated through a (jivev
range of temperature. Show that the increase in volume is the same
as it would be for a solid cylinder of the same dimensions composed of the
same substance and heated through the same range of temperature.
(t) Hollow cylinder
Let and be the radius and length of the cylinder at
and J?2 h radius and length at tjy. If a is the coefficient
of linear expansion, then we have
and +
Volume
But
and
r2=7r[i2.(l+a((2-/,)]* [/id +aC/,_^i)]
=7ri?i*/i[{l-f-a(^2-h)]*
<,)] (approx).
y=3a
^ 2 1 A-y{ti — h)]
Change in volume^Fa— Fi=Fiy(/2—
(«■) Solid Cylinder
The change in volume for the rise of temperature
IS given by ^
from to t^^C
This IS the same as for hollow cylinder calculated above.
is. Tfe volume 0/ the bulb of a mercury thermometer at
i!:&t , -4
of Tf 42. ^ dTid thr^ Cfihxcdlr^
U the lengtiof the L ^he L%C^Zl 4
T/ , ^1 • , {A*3£\l wE A
^o“volume of the bulb or th^i-
Wh„ .he b.,h U heated .e he.h ,£t,h" . “4T.
28
ENGINEERING PHYSICS
• •
or
it expand according to their coefficients of expansion.
Let T’6=Volume of the bulb at t°C
L,„=Volume of the mercury at t°C.
and T"m=l"o(l +r»«0
The volume of mercury that will be expelled out of the bulb
when heated to t°C. is
Vm- r6=Vo( 1 +ymO -Voi 1 +ygt)
—Voivin'-yg)^
Now if Jo=cross-section of the capillary tube at O^C its value
at rc.
If I is the length of the mercury column at t°C, than
where ?=2a^
I At—Voiym’—yg)^
[ym—yg] *
At
J_ ro(vm — yg)^
(1 -\-l%gt)
Sr clnelT^mar slre^-
thermal strains* , xv u
. ..SK r srM; isa
upon the coefficient of linear expansion. _
8 I=Exten 5 ion of the bar when free to expand-aTf
where ;=original length of the bar.
a=Coefficient of linear expansion
peevS Ty^e^thlr Vt^^ratlts^nd" thi of
• external compressive forces.
Extension prev ented
Thermal - strain produced=— original length ^
olTI
U
I
V
expansion op solids *
Thermal stress =Thermal strain x
= ExT
Heuce total Force =EixTA
where ^ = area of cross section of the bar and
£=young’s modulus of elasticity of the material
of tlie bar.
If the bodies are allowed to expand or contract freely witij the
rise or fall of temperature, no stresses will be developed in the body.
Also the contraction caused by lowering of temperature can
be checked by applying tensile forces to the bar resulting in thermal
stresses in it.
It should be remembered that (<) when a bar is heated and its
expansion is prevented the stress set up in the material of the bar
is compressive and (tt) when a bar is cooled and its contraction is
prevented the stress set up in the material of the bar is tensile.
Example 16. The rails of a tramivaij are welded together at lO^C.
Calculate the stresses produced in the rails when heated by sun to
Given that
Coefficient of linear expansion for stecl=^ \2'5 X
Young's modulus of sleel=2\00 Tonnes/cm-.
Si
Thermal strain -^ = a2'=l2'5 X I0-« X (45 — 10)
= 0*0004375
Thermal stress=£ x strain = 2100 x 0*0004375
= 0-91875 Tonoes/sq.cm.
Example 17. A steel wire 1 mm. in diameter is just stretched between
two fixed points at a iemperatare of 2o°C. Determine the tension when the
temperature falls to lo'^C. Given that
Coefficient of linear expansion for 0*00001
and young’s modulus for steel=2'14 X 10*^ k-gm/in^.
Thermal strain= — = . — nj'
I I
Thermal stress^^^.aT
Tension (Force) =F=E. A .aT kgm.
= 2.I4X10®-/4 (0.1)2X0.00001 1 X
(25—15)
= 1’82 kgm.
-30
ENGINEERING PHYSICS
Example 18- An underground pipe line is laid in spring at 15 C .
What stress irould be produced in it when the temperature falls doym
to—2^C in winter the pipe line is unable to contract.
of linear expansion of the pipe=0 00001‘ZI°C and £'==- X JO kgmjcm .
a.Tl
Tliermal strain
I
= ar
Thermal stress produced
=iy.TE
= 0 000012X[I5-(— 2)]X2X10«
=408 kgm-/cm~
Expected Questions
, , Define the cocflicient of hn^r expansion, ^s^^enorenln S'’ln"n^
determined. Mention a few cases where you meet this phenomenon in
praotice.
2. What happens to a sAid w^hen heat is applied to it. if
(ij It is isotropic
(u) it is anisotropic.
How is this eilect made use of in
a metallic thermostat
^6) scaling a wire in glass
tc) rivetting of boiler plates M.l.E.)
[it) a clock pendulum.
3 What is the relationship between the coefficient of linew expansion an
(t) superficial expansion (li) cubical expansion of a homogenous me . j
4. ^Vhat action is taken to ^ a/id^baU
track, gas or water mams, iron bridges, pendulums of cloc [A.M.l.E.)
lx AC
(A
5. Explain the working of toluene thermostat and vapour pressure operated
thermostat used in referigrators.
CHAPTER II
EXPANSION OF LIQUIDS
2.1. Expansion of liquids. Like solids, liquids when heated also
expand. As the liquids alwaj^s take the shape of the vessel in which
they are placed, it is quite meaningless to speak of their linear and
superficial expansions. It is the increase in volume which they suffer
Hence i'nfh. ^■’^Pansion.
-Hence in the case of liquids only cubical expansion is measured.
in f expansion. To heat a liquid it has to be contained
vessel is heated the containing
latter k L® simultaneously although the expansion of the
h cXd *! <=°'’®‘dered then the observed expansion
IS called the apparent expansion of the liquid. ansion
of apparent expjxmsion (y„) of a lium,l is the ratio
Apparent (observed) increase in volume Vt—Vo
Original Volume x rise in temperature ^
^t—Vo (l+yat)
real f “ «/
original volume of the liquid at'o'c. temperature to the
• •
al increase in volume
Original volume x rise in temperature
^<=l^o(l+yrO
Vt~~V,
Vot
31
ENGINEERING PHYSICS'.
32
where T'o and T', are the volumes of the liquid at and
fC and i is the rise in temperature above 0°C.
Note. The coefficient of real expansion (yr) is also called as the
coefficient of absolute expansion.
Hence in the case of liquids, two expansions, real and apparent
have to be considered. The real or true expansion of the liquid is
greater than the apparent expansion since it also takes into account
the expansion of the vessel.
Table for the coefficient of real expansion yr
Substance
1 -jV per ®C
! Substance
Yr per ®C
1
Alcohol
v-ouius
Oivel oil (
0.0007
Benzene
000114 ]
Paraffin oil
0.0009
Glycerine
Water
Mercury
000053
, 0 00015
000018
Pentane
Toluene
Turpentine oil
0.00159
0.00109
0.00095
2.4. Relation between Co-efficient of real and apparent expansions*
The real expansion of a liquid is always greater than the apparant
axpensioD. To show this experimentally.take a
glass flask with a long narrow neck and hll it
with coloured water upto the mark A. Now
suddenly put this flask into hot water
contained in a trough. It will be ob^rr-ed
that the level of the liquid falls in the first
instance to B and then slowly ”ses to C. This
is because of the fact that when the
immersed in hot water, it
from the hot water and expands.Therefore, the
levTl of the liquid falls to B. Thereafter he
liquid expands and the expansion of the
liquid also being greater than that
vessel its level ultimately rises to C from B.
In fact, we observe the expansion from A to
n“'volumrof lirportion measures the apparent expansion.
voC: .1 .h. p».io» BO .... ...I ..
°",w“ M.«. O. .h. por.to.
vessel for a given rise of temperature.
Hence Real expansion , ,^,,TT„oansion of the vessel (^B)
BC=Apparent expansion {AC) +Expansion 01
Let iind ra"
Fig. 2.1
EXPANSION OF LIQUIDS ^
0 \,
of the liquid, yg is the coefficient of cubical expansion of glass and
is the rise of temperature, then
Real expansion BC=yr XV Xt
Apparant expansion AG=ya xV xt
Expansion of vessel AB=ygX\’ XI
Substituting in (t),'we have
yr V Xf = ya Vxt + yg V Xt
yr — ya-^yg
or Coefficient of real expansion of a liquid
=Coefficient of apparent expansion of the liquid
^Coefficient of cubical expansion of the material of the vessel.
2 5. Effect of temperature on the density of a liquid. Whenever a
liquid IS heated, there is a change in the densitv of the liouid
Consider a mass ^ of a liquid and let and p, be the values of its
density at temperatures O^C and t^C respectively.
Since the mass remains constant, the volumes of the linuid V
and Vt at temperatures o°C and t^C, respectively are given by^ ®
and
But
where
Vo Pt
Vt=Vo{l+yr i)
yr= coefficient of real expansion of the liquid.
Pt
P8=P((l+yr /)
Determination of Coeffici
coefficient of apparent expansion o
any of the following methods.
of Apparent
liquid can
expansion (ya). The
be determined by
(•) The volume thermometer or Dilatometer metliod.
(ti) The Weight thermometer method.
(•») The hydrostatic or Sinker method.
Fig. 2.3.
?4 ENGINEERING PHYSICS
(i) The volume thermometer or dilatometer method. A dilatometer
consists of a bulb blown at the lower end of a
thermometer tubing as shown. The volume of
the bulb and the volume of each division of the
stem are first determined. The experimental
liquid is filled in the dilatometer after which it is
]ilaced in an ice bath. The initial volume (To)
the liquid at 0°C is found. The dilatometer is
then placed in a bath having a temperature fC.
When the level of the liquid, which falls in the
first instance and then rises up, becomes steady,
the final volume (T^) of the liquid is found.
Tf To and Vt are the volumes of the liquid
at 0°C and eC respectively then the increase in
volume V can be readily calculated from the
known volume corresponding to each division.
Thus the coefficient of apparant expansion is
given by
V^t
Incre ase in Volume
^Ori^inal volume x rise in temperature
If the initial tempeVrture is not but and the final
temperature is t^C then,
Increase in Vo lume
^Original volume xnse in temperatme
Where 7, is the volume at and n the volume of the same
mass of the liquid at f^C.
(ii) The Weight Thermometer method.
This offers a more accurate method for
measuring the coefficient of
pansion of a liquid. since it is based on the
determination of weight and not
It consists of an elongated glass "
silver bulb B provided with a bent
capillary tube C drawn into a narrow
iet at the end as shown m h g.
\ 4 In order to find the coefficient of
Apparent expansion tlm bulb >s cleaned
dried and weighed when empty. It is
then filled with the liquid “effici-
ent of apparent expansion is to be deter
mined. The filling is done by the method ^ ^
nf alternate heating and cooling of the ^ , • -j
ilb i Ld keeping the end of the bent capillary tube always ins.de the
expansion of liquids
35
liquid (to be filled) contained in a beaker A. When filled completely, it
is placed for some time in a water-bath at the room temperature,
keeping the end of the capillary tube still in the liquid in the
beaker. It is then removed and carefully dried, any drops of the liquid
adhering to the end of C are removed by blotting paper and is
weighed. The difference of the two masses is the mass of the liquid
filling the thermometer at the room temperature
Now the thermometer filled with the liquid is suspended in a
water bath in which water is kept at a constant higher temperature
The end of the capillary tube is kept outside the hot water
bath. As the liquid gets heated, it expands and is expelled out of
the thermometer. When no more of the liquid comes out, the liquid
has attained a constant temperature of the bath. The thermo-
meter is taken out of the bath, wiped well on its outside and allowed
to cool to the room temperature. Its weight is again taken. From
the three weighings thus made the coefficient of apparant expansion
can be found out as follows.
Let m=the mass of the empty weight thermometer,
77 ij=its mass when filled with liquid at the initial temperature
ti^C,
m2=its niass when filled with liquid at the final temperature
J/i^mi-m^mass of the liquid that fills the thermometer
at
'''Jtfa=m,-m=mass of the liquid that fills the thermometer
rise in temperature.
1 volumes of the weight thermometer and p,
and P2 the densities of the liquid at the initial and final temperatures
respectively, the coefficient of real expansion of the liquid and y«
Iher^o'^teT expansion of the material of the weight
From the definition of density, we have.
• •
Pi— ^ and p.,—
* 1
M
8
M
V,P
Now or =
and F,=K, (l+y,0or^ +
36
ENGINEERING PHYSICS
Substituting these values in equation (») we have
1 yrf
il /2 1 -\-ygt
Cross multiplj’ing and simplifying for yr, we have
yr= ;
3/, -M
3/2 t.
* 4 -
3 /,
3 /.
...{it)
A simple example will show that for many liquids it is permissi-
ble to put-^^’ =1.
j\l 2
From equation (u) we have
3/1-3/2 , .
ii'
3/1 -3/2
yr-y3= -jrj
But yr— y; 7 =/ci=coefficient of apparent expansion
Coefficient of apparent expansion yo=
JJ-^ t
Mass of the liquid expelled
“ Mass of liquid left^t higher temp, xrise of temp.
The weight thermometer thus enables, the coefficient of apparent
expansion of a liquid to be easily determined. If the coefficient of real
expansion of the liquid is required, the coefficient of cubical expan-
sion yn of the material of the thermometer must be added to the
coefficient of apparent expansion, but this method does not give a
very accurate value for the coefficient of real expansion as yj, is not
known to a high degree of accuracy.
If the’ weight thermometer is made of quartz or silica or p\Tcx,
the magnitude of yg would then be so small as to be negligible in
comparison with the coefficient of apparent expansion, so by ignoring
the expansion of the bulb t.e..by putting yy=0 in equation (n) we have
Cofficient of real expansion r
The advantages of using silica in place of glass for the bulb are
(t) It can withstand fluctuations in temperature.
{ti) Its coefficient of cubical expansion is smaller.
Since weight can be measured with great accuracy, tlie weight
thermometer is preferable to a volume method of measunng expansion.
It is, however, not suitable for use with volatile liquids.
EXPANSION OP LIQUIDS
37
Example l. A weight thermometer contains 101'7f} gim. of a liquid
at 14°C. On heating it to lOO^C, J‘75 gms. of the liquid are expelled. Find
the coefficient of absolute expansion of the liquid if the coefficient of
linear expansion of glass is 0 000009.
Mass of liquid at 14°C
Mass of liquid expelled (J/j —
Mass of liquid left behind
Rise of tempecature (/a — =
Coefficient of apparent expansion ya =
101 ‘75 gms.
= 1-75 gms.
iOl-75— 1-75 = 100 gms
100 — 14 = 86'’C'
1-75
100X86
= 0 000009
= 2 03 X 10-^
0 000009 X 3
: 0*000027
glass he
Coeff. of linear expansion of glass ot
Coeff. of cubical expansion of glass yg
Hence y^ — ya^yg
= 2 03X10-H0'27x 10^^
= 10-^[2*03*f 0*27]
= 2-3x 10-^ = 0 00023
Example^ 2 . If the coefficient of apparent expansion of rnercury in
1
y’hat mass of mercury will overflow fro 7 n a weight
thermometer which contains 400 am. of inprcur^, nt i .
perature is raised to 90^C. ^ ^ <ii 0 C when the tem^
Coefficient of apparent expansion ya=
1
6500
But
mass expelled
mass left at higher temp, xf,
mass expelled
(total mass— mass expelled) xt
Denoting the mass expelled by we have
1
X
or
or
6500 (400 — a:)X90
36000 — 90a:=6500a:
^=SS7 gms.
38
ENGINEERING PHYSICS
Example 3. If ft iveight thermometer is completely filled with 275
gm. of mercury at 0°C, calculate the mass of mercury which will fill it
at ISO^C. The coefficiejit of linear expansion of the material of the weight
thermometer being .9 x per'^C and coefficient of real expansion of
mercury is 182 Y. 10~^ per°C.
Co-efficient of linear expansion of the material = 9x 10"®.
Coeff. of cubical expansion y^=3 X 9 x 10“®.
Now y^^=lyJ.— yg
= 182 X 10-® — 3 X 9 X 10-®
= 155 X I0-® per V.
Let M 2 gm. be the mass of mercury which would fill the weight
thermometer at I00®C.
Mass expelled = (Total mass— il/ 2 ) = (274— 3 / 2 )
mass expelled 275—
Coeff. of app. expansion ya= Hias ' s lirt^^^J/^T^O
155 X 10"®=^
275 — T/
t
M 2 X 1 00
or 155XiV2X 10-‘ = 275— 3/2
or 3 / 2(1 + 1 55 X iO-^) =275
M = -
* 1 + 155x10-**
275
10155
= 269 2 gms
Example 4. 45 grams of alcohol are needed to completely fill up a
weight thermometer at 15^C. Calculate the weight of the alcohol which
will overflow when the weight thermometer is heated to 33^C. Given ya
for alcohol is 0’0012l.
Mass of alcohol completely filling
the weight thermometer at 15°C 3/i=45 gms.
Rise in temperature (=33— 15 = I8°C.
Let m be the mass of alcohol expelled
Mass of alcohol left behind 3/2=(45 — m) gm.
Hence
• •
m=0‘96 gms.
EXPANSION OF LIQUIDS
LEAO SHOTS
Fig. 2-5.
(m) The Hydrostatic or Sinker Method.
Besides the weight thermometer method, there is a hydrostatic
method or Matthissen’s method for
determining the coefficient of apparent hook
expansion which utilises Archi-
mode's principle.
A sinker which may be a cube
or a bulb of glass or silica as shown
in fig. 2.5 loaded with lead shots, so // ^
that it just sinks in the liquid, is ll SINKER
first weighed in air and then weighed (I j|
when it is totally immersed in the
liquid at room temp. The liquid
is then heated to a new constant '
temperature l2^C, and the new appa-
rent weight of the sinker in the liquid L£AO SHOTS
at is again measured. In order to
avoid inaccuracy in weighing, the Pig 2 S
liquid is arranged under the balance
case and the sinker is suspended from the balance by a fine wire
passing through a hole in the bottom of the case.
Suppose ”»<>. «h and m, are the respective weights of the sinker
in a.r in the liquid at and in the liquid at l°C. Now Archi-
medes principle states that the loss in weight of a body immersed
m a liquid li equal to the weight of the liquid displaced by the body
In the present case, ^ r j- tuc uuu^ .
temp"erItu^V»^~'"‘=‘°^* the initia
temp"eratur??7c”“’~"“='°*^ " =‘"her at the final
=(^2— final and initial temperature of liquid
.. -r; “I
and p,==
or _:^=ZiZl = Ii Pi
P,^, F, P,
or -j^i - 1+yrJ^
-^2 1 +yjt
40
ENGINEERING PHYSICS
Cross-multiplying and simplifying for yr, we have
Again puting^* = j
j}l 2
(approx.) we have
All the quantities in the expression for the coefficient of apparent
expansion have been measured in the experiment and so it can be
calculated, but we cannot find the coefficient of real expansion of the
liquid unless the cubical expansion of the material of the sinker is
known.
Example 5. A piece of glass which weighs 92 gms, in air is found
to weigh 51'6 gms. in a liquid at lO'^C. At 95^C its apparent weight in
the same liquid is 53‘9 gm. If the coefficient of cubical expansion of glass
is 24 X find the coefficient of absolute expansion of the liquid.
Loss of weight at lO^C, 3fi = 92 — 51'6=40'4 gm.
Loss of weight at 95°^, 3/2=92— 53’9=38-l gm.
Coefficient of cubical expansion of glass, yj=24X
3/,— 3/2 _ 40-4— 3 8 1
38-lX(95— 10)
=: =0*000715/°C.
38-1X85
yr—ya+yg .
= 0-000715 + 0-000024
= 0-000739
Example 6. A piece of solid whose coefficient of linear expiansion
is a, weighs W gm. in air. It iveighs Wi gm. in water at and W 2 9^-
in water at V"C. Determine the coefficient of absolute expansion of
water. {A.Jil.l.E.)
EXPANSION OF LIQUIDS
Loss of weight in water at O^C, — W —
Loss of weight in water at t°C, i.
Coefficient of apparent expansion
41
yr=ya-\-yg
_ Mx -M
t
+ 3a
cm
Example 7. A loaded glass bulb weighs 156-25 gms. in air, 57-5
y./^. when immersed in a liquid at 15*^C and 58-7 gm. when immersed
at o2°C. Calculate the mean coefficients of real expansion of the
liquid between and 52^C. Coefficient of linear expansion of
glass^O X lO-^l'^C:
JWi=T7—iri= 156-25 — 57-50
= 98-75 gms.
• 56-25— 58-57
= 97-68 gms.
Rise in temperature ^=52 — 15 = 37*C
Jtfi — 98-75—97-68
ya =
3/2 t
107
97-68 X37
= 29-6X 10-5
But
where
97-68X 37
yr— ya+yj
y^=3a— 9X I0-«X3 = 27X lO"®
^2-7X 10-5
y^==29-6X10-«+2-7 XlO-5
= 32-3X10-5 per ®C.
2 . 7 . Principle of U-tnbe for Comparing densities of liquids- In order
to compare the densities of two liquids
^yhich do not mix with each other, the
liquids are poured in the vertical limbs
AC and DB of a U-tube. Let the surface
of separation of the two liquids be at D.
if the height of the liquid of density Pj is
hx, and that of the other liquid of density
Pi IS Ag.above the horizontal line DC, then
Pressure at D—hxPxg-i-P
Pressure at C=h^Psg-\-P
where g is the acceleration due to
gravity and P is the atmospheric pressure.
Since the points C and Z) are in the
same horizontal line,
^Pi9A-P==hP2gA-P Pig ^
42
ENGINEERING PHYSICS
or
Thus the heights of the balancing liquid colu
proportional to their densities-
Milk:
are
inversely
2.8. Measurement of real or absolute expansion of liquids-
The co-ffficient of real expansion of a liquid can be determined
by the following methods.
(i) Dulong and Petit’s method.
(j») Regnault’s method.
fij Dulong and Petit's method. It is important to devise a way
of finding the coefficient of real expansion of a liquid which is
independent of the expansion of the containing vessel. A method
for determining the coefficient of real expansion of a liquid in whieh
expansion of the vessel has no effect on the observations, was
37SAM
COLD B
WATBP
Fig. 2-7.
j , Dnlone and Petit in 1820. The method is based on the
famous hvdfostatic principle that if tu-o liquid columns balance each
othlr their' heights are inversely proportional to their densities, this
relation is independent of the diameter of the tube.
The simple diagram of the apparatus is shou-n in the Fig- 2.7. It
• Vf ^Hass r tube ABCD, the limbs AB and CD are vertical
w^'iirthe porfion BC is horizontal. The vertical limbs AB and CD
^ bounded bv wider glass tubes with their ends closed with corks.
^ n^S^^both thJ upper and lower corks pass narrow bent tubes
the purpose of inlets or outlets. Two thermometers
T and^r! are also ins^ertS through the corks in the two Umbs as shown.
'Thl linilid whose coefficient of real expansion is to be determined
is fluid the u-tube so that its level is visible in each hmb.
EXPANSION OF LIQUIDS
43 --
One of the limbs, say AB is kept cold by circulating ice cold water in
the outer tube. The water enters at the lower inlet tube and goes
out through the upper outlet. The liquid in the limb is heated
by passing steam in the outer tube. The steam enters from the
upper end and leaves at the lower end. Liquid in both the limbs of
U-tube is the same, but owing to the difference in temperature, tlie
density of the liquid on the two sides is different. In order to
prevent the conduction of heat from the hot to the cold limb, the
horizontal portion BC of the tube is covered with cloth over which
cold water is constantly poured.
The two limbs and hence liquid in them is thus maintained at
different temperatures. The liquid in CD is heated to tlie temper-
ature of steam and its level rises. When the liquid level in tlie two
limbs becomes stationary and an equilibrium state has been reached,
the heights of the liquid level in AB and CD above the axis of the
horizontal tube BC are measured and the temperatures in the two
limbs are also noted.
Let the temperature of the liquid on the side AB=o'*C
Height the liquid column in AB ^ho
Temperature of the liquid on the side CD s=<®6’
Height of the liquid column in CD s^kt
Density of liquid at 0®<7
Density of liquid at
+ 1 , pressure exerted on the horizontal portion of the tube bv
the cold column is hoPo9 and that by the hot column is htPiO. As the
Uqurd columns are in equilibrium, the pressure exerted by them in
the same horizontal line is the same.
hoPog^hiPtg
hoPo=^lllP I
or
But
Pt
P^
Pt
fH
ho
it -hyrO where yr is the coefficient of
real expansion of the liquid.
ht
or
or
iIq
ho^hoyrt^ht
hoyft^hf — ho
hf "—ho
or yr^
hft t
The above expression is quite independent of the expansion
44
ENGINEERING PHYSICS
of the material of the containing vessel and hence gives us the
coefficient of real expansion of the liquid.
Dulong and Petit's experiment, though simple in design, was
subject to the following errors.
(/) A small portion of the liquid in each column has to project above
its constant temperature bath in order that its height may be read.
(it) Since surface tension varies with temperature its effect on
the two liquid surfaces will be different although this source of
error is eliminated to a great extent by making the upper ends of
the limbs wide.
{Hi) Since the two limbs are at a large distance apart, the small
difference in height {ht—ho) of the two columns cannot be accurately
measured, particulary in the case of mercury which has a small
coefficient of expansion.
(u') No efficient stiring arrangement can be used.
To overcome these errors, Regnault modified Dulong and
Petit’s original method and his apparatus is explained below.
(ii) Regnault's method. The apparatus consists of two pieces of glass
tubing ABCD and EFGH as shown in Fig. 2.8 joined by a flexible iron
tube to allow them to expand independently. The axes of the portions
BC of the tube ABCD and of the portion GF of the tube EFGH are
kept accurately in the same horizontal plane. The portions HG and
AB are enclosed in a wide bore tube. Similarly the portions FE and
CD are enclosed in wide bore tubes as shown.
The liquid whose coefficient of real or absolute expansion is to be
determined is filled into the tube so that it stands at the
Fig. 2.8.
same level in part HG and AB. The vertical arm CD is
EXPANSION OF LIQUIDS
45
surrounded with an oil bath and the portions EF, AB and HO
are jacketed with cold water baths fed from a common supply. The
liquid in the tube CD expands. The level in AB. therefore, rises
above the level in HG. When equilibrium is attained the various
heights indicated on the diagram and their corresponding tempera-
tures are measured. The portion DE is at room temperature CC. If
Pi, Pg and p are the densities of the liquid at the temperatures i.,
and CC respectively.
Liquid pressure on the side EFGB—}i^9^A-iixP\A-^P
Liquid pressure on the side ABCD=h^p^-\-
In the equUbrium position, we have
But Po=PiHA-Yrti) = p 2 (l-hYrti) = PilA-yr()
Where Po is the density of the liquid at 0®C and yr is the coefli-
cient of real or absolute expansion of the liquid.
^ equation we
^2Po ^ hiPo , HiPo , kPo
i^-f-Yrh) [l-i-YrtA (l+yrM (1+yrM {l+Yrt)
or — --1 Hi , h
(I+yrM (l+yrfal (* +yr^) { 1 +yr/i) "^(1+77)
in be‘lTsily‘cricukt?d. «<i"ation have been measured, so
i00°c rcSi *eh/ ’"'"“’•y. Umbs being at 0°C and
Zlum^Ts rrl' i ■'1 * and the hot
Height of the cold column Ao=60 cms.
Difference in the heights of the hot and cold column
ht — 4oal ‘80
Difference of temperatuie /=100— 0=I00®C
Let yr be the coefficient of absolute expansion of the liquid,
h~~K\ 1*8
yr—
ho t
60X100
0 003
Height of mercury column at 0 '>C=A <,=75 cms.
Height of mercury column at 100'>C=*, ^=76-35 cms.
Kise of temperature f=Ioo®C
46
ENGJNcERING PHYSICS
Now
h^xt
76-35—75
75X 100
= 0 00018
Example 10. The coefficient, of linear expansion of glass is
3x 10~^ afid coefficient of cubical expansion of mercury is l‘8xl0~^.
What volume of mercury ‘must he placed in the specific gravity bottle in
order that the volume of the bottle 7iot occupied by mercury shall be the
same at all temperatures ?
Let r,n and T'’,; be the volumes occupied ; by mercury and glass
respectively at 0°C. Let t°C be the rise in temperature.
Increase in volume of the glass when temperature rises through
t°C=VyXygXt
Increase in volume of the mercury when temperature rises through
t C=Vm X ym X t
The volume unoccupied by mercury will remain the same if the
increase in volume of the vessel for a given rise in temperature is
equal to the increase in volume of mercury for the same rise in
temperature.
or
Vgf X yg X i tn X ym X t
yg _ 24 X I 0-»_
2
or
Vg Ym 1-8X10-*
15
or
i.e., volume occupied by mercury must be
of the volume of
the glass vessel.
iTvamnle 11. The coeffideiit of real expansion of mercury is 0-00018
and the coefficient of linear expansion of glass is 0^000000. Find what
lenathofthe tube of glass 300 mm. long must be filled uuth mercury lu
order that the volume unoccupied by mercury may remain the same at
all temperatures.
Let Im be the length of the tube occupied by mercury and fC be
the rise in temperature.
Increase in volume of mercury when temperature rises by fC
=VmXymX^=a;yri
where a — area of cross-section of the tube, y,„ th^e coefficient of
real ^xpanslo';; of mercury and y, the coefficient of cubical expansion
of glass.
■ Increase in volume of the glass tube when the temperature rises
by W=L,y,f=a«.W*=«X300X 3X9X10-«Xf
EXPANSION OF LIQUIDS
47
For the volume of the glass tube unoccupied by mercury to
remain constant.
Increase in volume of mercury
= Increase in volume of the glass tube when the temperature rise
by t-^C
300 Ctygf-
or ?to=300 ^
ym
j 300X27X10-*
2.9. Aaomalous expansion of water. Almost all liquids
expand on being heated but water behaves in a peculiar
manner. When water at 0®C is heated its volume decreases and
300 X27
iTo
=45 mm.
increases upto a temperature of 4^C. After 4°C
the rise in temperature causes an increase in volume and consenuentUr
vfrTaTion of therefore, has a maximum density^t Jc
[Note. The volume of water is minimum at nr jfc a
temperature starts expanding though not'’''uniformir*'"lts''°c"™ff- ■“"‘1
of expansion .s different between "different "l^gTs^of tempe^atoef
other aqn\^ri“"::rt"o “ur:i;rl"w^tt%\m
the temperature falls below o°C and n, nrfJ ^ 90Untnes where
freeze. Let us. for expample co^ -nd seas
the water surface is below O^C* here the air above.
4S
ENGiNEEKiNG PHYSICS
As the atmospheric temperature comes down the upper layers of
water in lakes etc., cool and contract and sink to the bottom. This
continues till the temperature of water in the lake reaches 4°C. When
the top layers cool further, below the cold water being lighter
does not go down but as further cooling takes place the top layer
gradually freezes. Both water and ice are bad conductors of heat.
The lower layers are, therefore, protected from freezing. Thus we
see that water continues to e.xist at the bottom at though a
thick layer of ice may have formed at the top. Thus marine life is
saved from e.xtinction by this anomalous expansion of water.
If water had contracted regularly like other liquids, the colder
water would have settled at the bottom and ice formed at the bottom
of the lake instead of at the top. During the short span of cold
weather, the whole of the lake would have frozen that is would have
become solid ice, thus perishing the entire marine life.
Example- 12- piece of glass weighs 40 gtns in vacuum and 28'0
ijm. when immersed in water at 4°C and ■J8-4S6 gm. in water at its boiling
point. If the coefficient of cubical expansion at glass is 27 X 10~* per
'^C.find the density of water at 100°C.
Mass of water displaced by glass piece at 4®C.
7;j,^s=3it?— m‘j= 40*5 — 28 = 12*5 gms.
Mass of water displaced by glass piece at lOO^C.
7n2=it’—H’2=40-5— 28-486= 12-0 14 gm-
Rise in temperature= 100 — 4=96®C
^ = - — L11®11- = 42-I8 X 10-« per ‘’C
•• m.yXt 12-014X96
But yr=yci+y 3 = 42-18X I0-«-i-27 X IQ-*
= 69-18 X 10“® per '’C.
The specific volume of water is minimum at 4®C or its density is
maximum at 4 'C and is equal to 1 gm^ern®
P4=lgm/cm='
But P4=P.oo(l+TrO=Pio9{l+69 »8X10-«X96)
I=Pi00 (1+69-18X 10-8X96)
1
or Pioo® (H-69-18X 10-8X 96)
= {1 — 69 I8X 10 -«X 96 )= 1 — 0-0662 = 0*9338 gm/cm*
Expected Questions
I (a) Define coefficient of cubical expansion. What is the distinction between
real and apparent expansion in the case of a liquid ?
EXPANSION OF LIQUIDS
49
(b) Explain the terms absolute expansion and apparent expansion and how
they are related. (A.M.I.E.)
2. Explain proving any formula involved, how a dilatometer is used
tor measuring the coefficients of expansion of a liquid.
3. De^ribe with relevant theory, Regnault's methods of determining the
absolute co-efficient of expansion. j £
• 1 ^ a solid of known coefficient of expansion and which
sinks in the liquid. Describe giving relevant theory, how vo j would determine the
real coefficient of expansion of the liquid.
5. Write a note on the anomalous expansion of water.
(-■1. M. I. E.)
6. Water is not at all suitable as thcrinomotric liquid between 0®C and s®r
P''°Pe''ty of water saves the lives of the acquatic animals in very
Explain this satement as clearly as possible.
(.•\cro Society of Engineers]
Krx-K '
I (
{ • ^
\
\
\
V
A a ^«’ •
D.<» ^
'> i ,s .
CHAPTER III
EXPANSION OF GASES
3.1. Introduction. Jlechanical engineers who design engines of all
types must know how the gases inside them expand and contract
when they are subjected to changes of temperature and pressure.
O 1 this account the subject of “The Expansion of Gases has consi-
derable practical importance.
Gases like solids and liquids change in volume with change
in temperature and they do so at a much higher rate. Moreover.
\vhile discussing the expansion of solids the effect of
not taken into consideration because the change m volume of a solid
or a hquid with pressure is almost negligible. But in the case of
eases ^the change in pressure, considerably affects the volume even
fhen-the temperature is kept constant. Thus o a gas
we must take into account three variables— the volume (1 ).
temperature (T) and the pressure (F).
These variables are related to and mutually depend ^p^^
another and in order to study the relation between any two. the third
must be kept constant. We can have the following three relation-
ships ;
(i) the relation between pressure and volume at cons an
temperature (Boyle’s Law).
(»■) the relation between volume and temperature at constant
pressure (Charle's Law).
(in) the relation between pressure and temperature at constant
volume (Law of pressures or Gay Lussac s Law).
These laws are known as Gas Laws.
3 2 Bovle’s Law. Robert Boyle in 1 662 disco vered that at
temperature the volume of a given mass of a gas is inverse y propor i
to the pressure. ^
Mathematically Per. -y
pp=a constant ==.K^
50
EXPANSION OF GASES
51
The constant K depends upon the
mass, nature and the temperature
of the gas. The equation PT’=constant
is called the isothermal equation
of the state of a gas. If the pr^sure
and volume of a gas change from
Pg- ^^2 constant tempe-
rature, then the equation is
1\V,=P.V^= =K
If a curve is drawn for P and V
taking V along the abscissa and
P along the ordinate. then the curve will
be a rectangular hyperbola as shown in
the Fig. 3 . 1 .
Fig. 3.1.
Limitations of Boyle's Law. Boyle verified this law for a small
range of pressures. Despretz showed that the volumes of carbon
dioxide and ammonia decreased more rapidly with increase of
M Boyle’s law. Hence this law
does not hold good rigorously at high pressures.
Further investigations showed that it is true for any gas remote
from Its point of liquefaction. Hence this law is correft for the
ratures remote from their temperatures of liquefaction ^
According to Boyle’s law
p,r,=p,r3
IX260=:P2X150
P3 = 2«».
150
= 1‘733 atmospheres
leiigth of at. metcury\llumn^^{s when in fnercury the
the volume of the space above mercum \ l depressed so
the mercury column is only T 2 cms. Find t?e
no changrin" the^tngthTfTe ^mer^^T?
place when the tube is depressed As should take
depressing the tube, some air is present ab^": ih?^::rcL^tS
pressures exerted by ^thT^S^ presfuribwe
cases, ^ present above mercury in the two
Apparent atmospheric pressure in case 1=74-0 cm of Hg.
52
ENGINEERING PHYSICS
P=74 + Pi or Fi=(P— 74)
Apparent atmospiieric pressure in case IT=72 cm of Hg.
P=12 + P2 P2—{P — 72 )
Let T’i and l'.> be the volumes of the air persent in the space
above mercury in the two cases respectively, then
According to Boyle’s Law.
74) 72) ]\
or (/*— 74)I'i = (F~72)I'i;'2
or 27^— 148=/*— 72
or /*=148— 72
= 76-0cmsof Hg.
Example. 3. The volume of Torricellian Vacuum is Gcm^ ami its
length 12 cm. If 7 cm^ of air at the pressure of the atmosphere, which is
70 cm. are introductd into the barometer lube, by hou' much will the
the mercury column hi ihpiessul ?
The volume of air at atmospheric pressure before being,
introduced in the barometer tube I’l— 7 cm®.
Height of mercury column=Atmospheric
pressure. 7*1=76 cm.
Let this air depress the mercury column by .r cm.
Increase in volume of air=a*x — =0’5 x
The volume of air in the tube I 2 =(6-}-0*5.t*) cm®
Pressure of air 7*2=ar cm of mercury.
By Boyle’s Law, we have
P^\\=P^V._
76 X 7=a:x (6-f 0-5 .r)
532 = 6a:H-0*5.r2
or 1064 = 0
6ih-v/(6)®-«- 4X 1064
x=
2X6
6i:V^_ 6±65 5
12
12
= 5*96 cm.
Example. 4* A little air has leaked into a barometer tube one
metre lonq. The merew'y stands at 75 ems' mark, when the tube is vertical
and at 83 cm. mark wkeri the tube is inclined at 30° to the vertical. What
is atmospheric pressure 1
EXPANSION OF GASES
53
Let the true atmospheric pressure be P cm.
of mercury column.
(») Whem the tube is vertical.
Pressure of air inside the tube
Pi=(P — 75) cm. of mercury
and volum of air T\ = (l00 — 75) a=25 a cm®
where a=area of cross-section of the tube.
(u) When the tube is inclined
A ertical height of mercury cohimn = 83 xcos 30®
= 83 XO-866
_ =71*9 cm.
Pressure of air inside the tube
P 2 ~{P — 71*9) cm. of mercury
and volume of air (100 — 83) a=l7o cm.®
By Boyle's Law, we have
(P- 75) X 2 5a= {P~ 7 1 -9) X J 7o
(P-75)X25=(P-71*9) J7
25P-I875 = I7P-1222*3
8P= 1875— 1222*3 = 652-7
I'ig. 3.1.
or
P=
652*7
8
= 81*6 cm. of Hg
raturtof Ch
states that-P,.eaa«. co,.,ant. tke^olurnelf
Oas increase, ior, ecrea.es) t.yae constant fracUon^^ of .oW a*
0°C for each degree centigrade rise (or fall) in temperature.
The fraction _ (0-00367) is the coefficient of increase of
yo^me at constant pressure for all gases 7 / *
(I +yp<)
Vt-V,
0
ii)
H
54
ENGINEERING PHYSICS
The value of yp depends upon the scale of temperature used.
and = or ( 0 - 002 1 7) per
The equation (/) can be written as
( I +,73 j '''hen f is in
( 1 +4^o' )
Example- 5. In an experiment for finding the coefficient cf expansion
of air at constant pressure the volume of air at 30^C was 75 c c. ^\ hen the
air was heated to US'^C, ke.ejnyig the pressure constant, it occupied 92 c.c.
Calculate the coefficient of expansion of air.
Let \\ be the volume of air at 30°C and 9%°C.
Let yp be the coeff. of expansion of air when pressure is kept
constant.
Hence applying the formula.
Now r/=T'o {\-^ypt)
where yp is the co-efficient of expansion of air at constant pressure
... ,'^ = r„(l+yp30) - W
and r 2 =r„ (l+yp98)
Dividing (ii) by (i) we have
l+98y7,_^
IH-Sbyp 75
or yp=0 00426
^ T T aw When a given mass of gas is heated, keeping
^ 3.4. Gay Lussac s Law^ ZressuvT increases. This is because the gas
Its volume cons , ^pand. But it is not allowed to do so, as the
on heating tends t P . ^j^gj-gfore, results in an increase of
volume is_ep . relation between the temperature and the
a gas at constant volume and this relation is known as
th^Law of pressures or constant volume law. It states that.
Tke voJu,.e of t r"/-” f O^C.treacTZ^/e
a qas iwreases (or decreases) by , oj its pressure m j
conliyrade rise (or fall) iu temperature.
This law is sometimes called as Second Charle’s law of pressure. ^
Thus if P„ and P, are the pressures of a gas at 0 °C and PC
respectively at constant volume, we have.
EXPANSION OF GASES
55
or
Pt=P^ (l + y«0
Pt-Pn
y«
where y,. is the coefficient of increase of pressure at constant
volume (or simply pressure coefficient^ and its value is or 0 00367
The. pressure coe fficient (y,,) is defined as the increase, in pressure jter
unit pressure at O^C per °C rise of temperature.
The value of y^ depends up on the scale of temperature used.
Thus
y« = 97 s or (0-00367) per
andyy=--^ or (0.00217) per ^P
460
Again relation (i) can be written as
P,=P„(l4- if < is in T
If the temperature t is measured in '"F, then
3.5. Absolute Temperature, If a given mass of a gas is slowly
cooled at constant pressure, its volume decreases uniformly with the
temperature. As the gas is below O'C its volume shrinks further and
theoretically there is a definite temperature at which the volume of
the gas would become zero. The temperature at which the volume of
the gas becomes zero is called the ateolute zero and represents the
lowest temperature that can be conceived, or attained.
Consider the relation r(=Fo^l + — - ^
F(=o, then
0 = 7 . (,+ -^)
If
or
^= — 273
;= the absolute zero is-273°C. In other words oV
IS 273 degrees above the absolute zero, 100°C is 373 decree's +i
orTM = (273+0
56
ENGINEERING PHYSICS
Thus 0®r=^:73^A' = 273®.-J
The value of absolute zero on Fahrenheit scale is— 460®T’.
Hence the absolute temperature corrsponding to r F is given by
r°i?=.(460-h/)
where R denotes Rankin's scale corresponding to absolute zero
of — 460®F.
3.6. Alternative Form of Charle’s Law. It states that tJu volvme of a
qivfn mass of a ga^ is proportional to its absolute temperature at constant
jtressure.
If I'l and W be the volumes of a given mass of a gas at t^^C and
tf^C respectively at constant pressure, then
273+< i
273
where ro“Volume of the gas at 0®C
T,^K={273-th°C)
and To^K^{n 2 -\-O^C)
V -r
* I — * 0 rn
0
or
Similarly
Ll=Zi
I ’o n
To ~T^
From {i) {^*)» have
r, -T,
= a constant
EXPANSION OP GASES
— constant
57
or
Hence T'^oc T
In other words the volume of a ^iven miss of a gas is proport-
ional to the absolute temperature, when pressure remains constant
Example- 6. A gas at lias its temperature raised so that its
volume is doubled, the pressure remaining constant. What is its final
temperature.
Let t°C be the temperature to which the gas has been lieated so
that its volume is doubled.
Initial temperature 7'i=273-|-l 3 = 286®.fl
Initial volume =1’!
Final volume =21'!
Final temperature A
As pressure remains constant, so according to Charle’s Law.
or r, =1-7^ X r,
^ 1
273-ff = ^!-^X286
* 1
or 273-i-f = 572
or f = 572 — 273
= 299°C
Example. 7. The density of Argon is 1-6 gmjlitre at 26°C and at a
prmwreo/76 cm of mercury. What is the mass of Argon in the gas
filled lamp bulb of volume 100 c.c., the pressure inside is 76 cm. of
mercury when the average temperature of the gas is 120'^C ?
Let voW of Argon at 120"C=(273 + I20)°^ and at a pressure
of 76cmof Hg.Fj^lOOc. c.
and the volume
pressure^Fj
at temperature 26^0, (273+26)®A', at the
According to Charle's law as the pressure remains constant
Ti-T,
or
^2=f,x ^ =ioox
299
same
393
58
ENGINEERING PHYSICS
I' =76 cm®.
Since the density of Argon at 26'’C — ]-5 gm/litre
• •
and
mars of 1 litre (lOOO cm®) of Argon at 26^C'=l-5 gm
mass of 76 cm® of Argon gas at 26*^0
1*5
1000
X 76 = 0*114 gm.
3.7. Alternative Form of Pressure Law or Gay Lussac's Law. It states
that at ronstaut volume the pressure exerted by a given mass of a gas is
jiroportiounl to its absolute temperature.
Let Pq and Pt be the pressures of a given mass of a gas at Q°C
and PC respectively at constant volume, then
or
or
where T and are the absolute temperatures coresponding to PC
and 0®r respective!}'.
In other words at constant volume, the pressure of a given mass
of gas is proportional to the absolute temperature.
If P and Pg pressures of a gas at absolute tempreaturea
T, and t\ respectively, then
^ iTi
59 -
expansion of gases
To
3.8. Relation between and yp- We have already stated
that experiment shows that the volume coefficient of a perfect gas
5 — — . the same value as for the piessure coefficients yp for a perfect
273 ’ *
and
or
gas. The equalit 3 ^ of yp and yy can also be shown to follow if the
gas obeys Boyle’s Law and Charle’s Law.
Let a given mass of a gas having a volume T’^o and pressure
O^C be heated twice from O'^C to t^C firsth' at constant volume (I o)
tc
Fig 3.4.
when the pressure changes from Pq to Pt and s^condlj’' at constant
pressure (Pg) when the volume changes from Vq to 1'^. Since the
temperature is the same in either case, it follows from Bo\de’s
Law that
But
PgF,=P(ro
EL=yL
Po
assuming Charle’s Law
Yl
t
273
Substituting the value of
in (l), we have
•60
ENGINEERING PHYSICS
Thus P,=P„(,+^'^)
It follows from the above relation that the pressure coefficient
7 » is also — .
^ 273
Thus the volume coefficient of the gas is equal to its pressure
coefficient if it obeys Boyle’s and Charle’s Law.
3.9. Ideal gas. An ideal gas or perfect gas is one which strictly obeys
tile eas law^* rtf\ croc .^ii
the gas laws. There is, however, no gas which perfectly obeys all the
gas laws. Some gases like hydrogen, nitrogen, oxygen, helium etc., the
so called permanent gases, obey the gas laws with sufficient accuracy.
These gases therefore, are considered to be perfect for all practical
purposes.
Example. 8. A given volume of air has 740 mm. pressure at 17**C.
What is the temperature in centigrade scale when its pressure is 1850 mm.
Initial pressure of air Pj=740 mm. = 74*0 cm.
Initial temperature of air Ti = 273+ 1 7=290®.4.
Final pressure of air P,— 1850 mm. = 185-0 cm.
Let PC be the final temperature of the air, then
Final temperature of air r 2 = (273+0°^-
.\pplying Gay Lussac’s Law, we have
^ A
- T.
Po
7*2= X 2 \
1850
74*0
_ X 290 = 725®A
But T=in+t
273H-/=725
or /= 452 °C
Example- 9. The volume of a gas at N-T.P- m 250 c.c. Its
Xemperaiure is increased to 5o^C and the pressure, by 145 mm., the volume
remaining constant. Find the pressure coefficient of the gas VJi
temperature.
Let P^ = 76.0 cm. be the normal pressure at 0°C or 273°--l and P 2
the pressure at 55®C.
P 2 = 76+14-5 = 90*5 cm.
and r 2 = 55°C=273 + 55=325“A.
Now Pi=Fo(l+rt-0
and
Now
9 0-5—7.5.0
76X55
14-5
76X55
= 0-00348
EXPANSION OF GASES
Example- 10. It is found that the volume of a certain gas inercaHcs
in the ratio 1‘035 \ 1 he.iimen 25°C and Calculate the ahsoJufe zero
on the centigrade scale for this gas.
Suppose the absolute zero is x^C Then assuming tliat the
change of volume takes place at constant pressure, we have
for this gas.
where and Ti are the absolute
1-035 .r + 25
“ x-f ] 5
temperatures
270-7®C
3.10. Gas Equation. The gas laws, discussed in tlie proceeding article-
give relations betw’een two variables of a gas when the third variable
is kept constant. If. however, all the three variables P.V and 7’, change
simultaneously, then the relationship between them is given bv an
equation known as Gas Equation wliich is a combination of Boyle’s
and Charle's Laws. ^
Let Vi be the volume of a given mass of a gas at a pressure I\
and temperature Tf'A. If the pressure changes to and tempera-
ture to Ti. the new volume ['g can be calculated as follows.
Suppose the whole process takes place in two steps In the
first step the pressure changes to P.^ but the temperature remains
constant. In such a case Boyle’s law is applicable. If. therefore
thenew volume is v, then
^i^i=^2^’=constant
K (. }u step suppose the pressure remains constant at 1\
“ = — constant (2)
T.
Hence substituting the value of t- from (i) in (2), we get
P V P V ^ o
= con£lant(say K)
This IS called the gas equation and in general is written as
^ =a constant
pressure P and the volume Y of a given
relation oc directly as its absolute temperature T Thp
wlfu the'vaTutr f- a gas
and T ^ f the three quantities P V
of constant I''ieptnds''‘luor^th^
concerned. '^Pon the mass and properties ot the gas
•62
ENGI^E6RING PHYSfCS
If V is the gram molecular volume, then constant K is
taken as R and is known as gram molecular gas constant or uni-
versal gas constant. Its value is the same for all gases because one
gram molecule of all gases at X.T.P. or S.T.P. (normal temperature
273®.-! and pressure 760 mm) has the same volume.
The equation then becomes
PV^RT
This equation is known as. characteristic gas equation.
In the C.G.S. system R is measured in erg/gni per
or cals/gm per
In the M.K.S. system R is measured in Joules/kg. per °A or
Kilo cal/kg per ^ A
In the F. P. S. system R is measured in ft. Ibs/lb per ® R
or B.Th. U/lb per ° R.
If we choose to deal with 1 gm. of a given gas, then the constant K
in equation is usually written as r and is called as the "Characteristic
Gas Constant" or simply the Gas constant.
Thus PV=rT for unit mass of a gas and this equation is known
as the characteristic equatioii of a perfect gas.
The value of r depends upon the nature of gas and hence its value
varies from gas to gas. The value of r for any gas can be calculated
from the above equation if the volume of the gas at X.T.P. or S.T. P.
is known. The units of r in different system of units are the same as
^^^*^^The gas constant r for oxygen can be calculated from the observa-
tions that the volume of 1 gm of the gas at X.T.P. is 700 c.c., the
pressure P of the gas is 76 cm. of mercury and the temperature of the
gas is 0®C so that its absolute temperature T is 273®--!.
h9gy.V
Hence
P\
T ~~ T
(76Xl3-6 X980)x700
273
Xote (0
= 2-6 X 10® ergs/gm/ degree A
P=shpg where h is the height of the mercury
column in cm.
Kr=sVolume in c.c.
y— Temperature in degrees absolute.
P is the density of mercury.
It* g is the acceleration due to gravity in cms/sec^.
We value of r in the case of hydrogen can be calculated as under.
ifpm of hydrogen occupies a volume of 1 1-2 litres at
1000 c.c., pressure P=76 cm. and Temperature T~112 A.
PV 76 X 13-6X 980X 11200 ergs/gm/M.
T~'^ 273
(u)
{Hi)
63
expansion of gases
Example. 11- 1 Kg. of air at K.T.P. occvpies 0 7734 cubic metre.
Calculate the value of the gas constant Jor the gas.
In M.K.S. system of units the pressure P is measured in
Newtons/metre*
/. P in Newton s/m^ = X 10'*
6 X = 1014 X 10® Newtons/m-.
“ 10 10
PT^
Applying the equation
we have
r= joulcs/Kg/M.
3 . 11 . General gas equation. We have seen that gas equation holds
good for unit mass of a gas. If now m gm. of the gas ha\ing a volume
V is considered, then the gas equation becomes
PV ^ mrT
or
This equation can be used to find the mass of certain a volume
of the gas, provided the gas constant r for one gm. is known. Further,
once the mass is calculated, the density p of the gas can easily be
determined from the relation p .
3.12. Universal gas Constant. If a gramme molecule of a gas is
taken the value of the constant K in the gas equation is denoted by R
and is known as gram-molecular gas constant or universal gas constant.
Its value is the same for all gases because one gramme molecule of all
gases at N.T.P. has the same volume of 22 4 litres.
The gram molecular gas equation is PV^RT
The gas equation can also be expressed by PV^=MrT where M
is the gram-malecular weight and r is the gas constant for one gram.
Hence Mxt=R i.e. M=-^—
r
Molecular weight ofagas= Universal gas constant
characteristic gas constant
Numerical Value of the Universal gas Constant.
The universal gas constant is given by the relation
^ T
Normal pressure =76 cm of mercury column
Temperatur e = 2 7 3 ®
64
ENGINEERING PHYSICS
Volume of one gm mol of gas = 22-4 litrs=22400 c.c.
76X 13.6X981 X22400
273
= 8*318 X 10" ergs per
gm-mol per
Since 1 Calorie
= 8-318 joules per gm-mol per °A.
=4-2 joules
li= -^--—=2 cals gm mole/°^
In the M.K.S. system^ Normal pressure
= 1.014 X 10^
7’ = 273M
(approx)
Newtons/m-
X’olunu* of om- Kilo — Mol. of gas = 22'4
m
H =
1.014 X I0’X22.4
273
- = 8310 joules; KiIo-mol/®.£4
as one
= 2A'-cal Kilo-mo]/®>4
Kilo— cal = 4200 joules
If till* pressure is expressed in kg m- instead of Newtons. «r, the
value of li would be
9.81
Gram Molecular :weight of a substance is the molecular wt. expressed in
crams. Similarly kilo-gram molecular wt. and Pound Molecular weight are mole-
cular wts. expressed in kgms and pounds respectively. The above quantities are
some times called as gm-mol. the K-mol—Jb-mol.
Some times the Universal gas equation is expressed in terms of.
Avagadro's Number (A) and Boltzmann constant (t).
3 13 Avogadro’s Number (N). The number of molecules in one gm
mole of a gas is called the Avagadro’s number. The number is constant
for all gases and its value is
^’■ = 6*02 X 10^^ per!77ft-mo/.
= 6*02X 10 ^® per A:— ?rto/.
3 14 Boltzmann Constant (k). The ratio of universal gas constant ^
and Avagadro’s number xY is called as the Boltzmann constant (k). Its
numerical value is
1:=^ 1 *38 X 10“^® ergs per degree/molecule
= 1 - 38 XI 0 “®^ joules/degree per molecule.
From universal gas equation we have
II
PV=IiT=^\
PV^NkT
where k
A
'n
expansion of gases
65
or
N
=kT
or
Pv=kT
where v=Volume of one molecule of a gas.
temperature Tj. Now from gas equation we have
Tr Tt
As density of a gas at a given pressure and temperature varies
invM^sely as its volume, then,
1 ^ ^
p^oc^and
y 1 2
/
Substituting the values in the gas equation, we have
= 3-
r,Pi TiP2.
if then, from above equation
Pz Pi
Hence the dencily of a gas at constant pressure varies invesely as the
absolute temperature.
X{ then, from the above equation
Pi Pt
Pi
<7
Hence the density of a gas at constant temperature varies
directly as the pressure.
Example. 13. 10 lbs, of air under a pressure of 1000 Ibsjsq. inch
has a temperature of 140^F. Find the volume occupied. Given
r=5Z-Z7 ft, Ih. per for 1 lb. of air,
P=s:1000Xl44 Ibs/sQ. ft.
T==4604-140 = 600®ii
w=10 lbs.
r=53’37 ft. lb. per **F per lb.
y— ?
66
ENGlNEERtNG PHYSICS
Applying, the equation
PV^mrT we have
1000 X 144 xr= 10 X 53-37 X 600
y_ 10X 53-37X600
“ 1000 X 144
= 2-224 C. ft.
Example. 14. .^4 litre of air at 0°C and under atmospheric pressure
weighs 1''2 gm. Find the mass of air required to produce at 1S°C a
pressure of 3 atmospheres in a volume of 75 c.c. M. M. /. E.)
Case I Pi=l atmosphere = l 013 X 10 « dynes/cm.2
Ty^ = 21 l°A, Fi = I litre = 1000 c.c.
m=\ 2 gm.
Applying equation P^\=mprTi
1*01 3 X 10® X 1000 = 1-2 XrX 273
1-013X10®X 1000 ^3.^^^^q,
r2X 273
Case IT P^ — 'i atmospheres = 3 X 1-0I.3X 10® dynes/cm®
7*2=— 18 + 273 = 255°^
T’’2=75 c.c. and m^—t
Again the equation P^\=m{rT^
_P^\ 3X1013X10®X75
or ^2 — 7 ^— 3-I2X10®X255
2
= 0-288 gm.
Examole. 15. 30 cm^ of Hydrogen were collected in a tube over
mercury at 28^C when the barometer stood at 75-8 cm If the mercury
inside the tube was 2 cm. higherlthan outside, calculate the volume of
the gas at N.T.P. (/I. Ji. i.
p^_7Jie pressure of Hydrogen gas collected over mercury
^ 75-8 — 2=73-8 cm. of mercury
Pj=Volume of the gas = 30 cm.^
y^=;Temperature=273 + 28 = 301°A
P,=Pressure at N.T.P. = 76 cm. of Mercury
r 2 =Temperature at N.T.P. = 273°A
^ 2 = Volume of the gas at N.T.P.='?
Applying gas equation, we have
P^Vl P^\
P<PANSION OF GASES
67
P,F, T2 _73*8 X30 273
Ti P; —301 76
^26 5 cm.®
Example. 16. A Hire of dry air weighs 1‘293 gms, at N.T.P., Find
the temperature at which a litre of air will weigh 1 gm., when the pressure
is 72 cm.
Initial volume of air Fi— 1 litre=i000 c.c.
Initial temperature T-i^li'i^A
Initial pressure P, = 76 cm. of mercury
Let P3 ’^A be the temperature at which 1 litre of air will weigh
1 gm.
Volume of I gm. of air =1 Iitre=l000 c.c.
Hence Volume of 1-293 gms. of air at r°.4=l000x l*293 = 1293c.c
^2 = I 293 c.c.
p2=72 cm. of mercury
Using General gas equation, we have
or
PiVi _-P2F,
‘^2
ri
76
>293
1000
273 = 335-4®^
= 335-4 — 273=62*4 ®C
Example. 18. Find the mass of a litre of a 7noist air at a temperature
of 32 C and a pressure of 758-2 m.m., the dewpoint being 15^C The
saturation pressure of aqueous vapour at 32^0 is 12-7 m.m. The densihi
of dry air at N.T.P. is 1-293 gmjlitre. [A.M IE)
According to Dalton’s Law of partial pressures, the total pressure
IS due to both air and water vapour,
/. Pressure exerted by air alone Pi=758*2— 12*7
=745*5 m.m.
Volume at 32®C and at pressure 745*5 m.m.
^1=1 litre
Temperature 27 3+32 = 305®^
Volume at N.T.P,
Normal Temp. rji=273®.4
Normal pressure Pj^yeo m.m.
Applying General gas equation, we have
68
ENGINEERING PHYSICS
PiVi_P2y
3
745-5 X 1
T
3
760 xT.
305
or
V2 =
273
745-5 ^273
■ ■ — X
760 305
=0-878 litre
Density of air at iV.r.F. = I •?93 gm./litre
Mass of this volume of air at N.T.P .
= 0-878 X 1-293
= 1-136 gm.
Example- 18- Two glass bulbs A and B of 400 c.c. and
connected by a narrow tube of negligible volume The apparatus
air at 0°C and 76 cm. of Hg. pressure and sealed off. If the temperature
"if the glass bulb A is nJ raised to lOO^C. Find ii) the ne^a press^e
I the system (ii) the mass of air mUch is transferred from one bulb
to the other during heating. Take density of atr at h.T.F. as I
gmjlitre.
The mass of the air in the two bulbs remains constant before and
after heating.
Before heating.
Mass in bulb
From the equation PV=mRT, we have
FaVa __ 76X400
A, ma—
Mass in
bulb B, mb=
RTa 273 XF
76X200
RTb
Total mass m =ma+*^b—
273 XR
76 X 600
W*
273XR.
After heating. When the temperature of the bulb is raised to
lOO'^C, the new pressure in both the bulbs becomes P.
PX400
Mass in bulb .4 , m a—
, _ Fx 200
Mass in bulb B, ni b 273 xR
, P ["400 («V
Total mass in both bulbs m ma'+mi, - J'"
As total mass before and after heating remains constant.
76X600 Pr400 ,
273 F R[_ 37 ^ 273 J
expansion of gases
69
_ P~f4OOx273 + 200 X373) ~|
373 X273 J
76X600X 373
^“lOO (1092 + 746) ■
^ 76X6X373 Qf jjg
1878
(ii) , For bulb A
fna
16>c^ at N.T.P.
273 Xi?
92-6X400
m a — -
373 Xi?
- at
100*C.
* * Wc 76 373
\t \ T P \ litre means 1000 c.c. and 1000 c.c. of air have mass
1*29 _ 1‘29
of 1-29 gm., hence 400 c.c. have mass of “ 2~
= 0-515 gms.
ma= 0-515 gms.
and Wa'= 0*89 Ttla
mass transferred = ma— 0-89 ma
= 0-515— 0*89X0*515
= 0*055 gra.
3 . 16 . Work done by a gas duringl expansion. Consider a certain
quantity of a gas enclosed in a cylinder fitted with a moveable and
frictionless piston of area A as shoNvn. Let the initial volume of the
gas be Vi. Suppose the gas is heated then both its temperature and
A 6
1
T T " '1
1 9 -
I 1
1
1 1
1
1 1
\ ;
1 p
1
• “I
1 t
1
1 ^
1
i :
1
? 1
j ^
Fig. 3.5.
volume wiU increase but any tendency for the pressure to increase will
PRESSURE
ENGINEERING PHYSICS
be counter- balanced by the outward movement of the piston. Hence
during tlie process of heating the pressure of the gas remains constant
and equal to tiie external pressure P and gas expands to volume V^.
Let during the expension of tlie gas the piston moves out through
a small distance dx.
Then work done by the gas
(/ir = Force acting on the piston x distance moved
= {Pressure (P) x (Area of the piston) xdx
~Px Ax dx
=P X (.'lr7.r)
But .<^<-7.r=increase in voIiime=f7T'
Work done, dW =PdV
If the volume of tlie gas changes by a finite amount from Fj to
then
Total work done, 1F=P (Fj— 1\)
(f) If P is in Xts 'irn} and dV in w®, then work done is in Joules-
(ft) If P is in dynes/cm.® and dV in cm.® then work done is in
Ergs.
When the gas expands as in this case the work is said to be done
by the gas (+i’e work done) whereas if the gas is allowed to compress
the work is said to be done on the gas {—vc work done).
The expansion or compression of a gas can be represented on a
graph drawn between P and I", pressure P is taken along the ordinate
and the volume 1' is taken along the abscissa. This diagram is known
as P— F diagram or Indicator diagram.
When the pressure is constant, the P — F diagram is the straight
line indicated by aa' as shown in Fig. 3.7 where constant pressure
is equal to AD or BC. If the volume of the gas changes from a value
represented by OD to a value represented by OC then
Change in volume=OC—
Thus work done
=Px change in volume
-Px(Fz-Fi)
=Area ABCD
Hence the ^vork done is given by the
shaded area bettt'ee>t the expansion line
and volume axis.
Now let the gas expand on heating
according to the PF curve AB from
an initial volume \\ to a final volume
Fig. 0.0
71
EXPANSION OF GASES
Fa corresponding to positions A and B of the piston. Let P be tlic
pressure and V the volume of the gas for any position of the piston .r
from the bottom of the cylinder. Although the pressure changes from
a value represented by KF to a value HO. when the volume increases
from OK to OH, we can imagine the pressure to be kept constant while
a very small change of volume dl’
Then the work done by the
gas is given by
(lW=PdV
t=work done by the gas in
e.Kpansion from OK to OH
=Shaded Area of KFGH
Total work done by the
gas when its volume changes
from Fj to T'g
w= j PdV
Fi ► VOLUME
=Total area under the curve AB. l ig. 3.7
We can thus formulate a general statement :
The work done by a gas is equal to the area between its P — diagram-
and the volume axis, taken between the limits corresponding to its initial
and final volumes.
Example* 19- A cylinder contains 3 litres of air at 2 atmospheric
pressure awd at 300^ A. The air is carried through the following operations,
[a) heated at constant pressure to 500° A {b) cooled at constant volume
to 2oO°K (c) cooled at constant pressure to 130° A. (d) healed at constant
volume to 300°A. Show each process in -pressure volume diagram giving
the numerical values of P and V at the end of each process. Calculate the
net work done. [A.M.I.E.)
Let at the initial condition of the air atraospheres,Fi = 3
litres and Ti=300®A) be represented by A in Fig. 3 . 6 . Let us consider
each operation separately.
IS made.
(a) Line AB represents heating of air at constant pressure. Since
pressure remains constant, applying Charle’s law we get
" 300
X3=s liters.
72
ENGINEERING PHYSICS
(fi) Line BC represents cooling at constant volume (Fig. 3.7).
Hence applying Gay— Lussac's Law, we have
£3
T.
or
3
^2
P,= ~lxP,=
250
500
X 2 = 1 atm.
(0
, Line CD represents cooling at constant pressure(Fig. 3.7).
Hence applying Charle's Law again, we have
T.
T
150
litres.
This volume is equal to the original volume V^.
((?) Line DA represents heating at constant volume (Fig. 3.7). It is
obvious that the air returns to the initial condition after this process.
Now the net work done by air in the whole operation
= Area enclosed by the shaded rectangle ABCD.
=ABxAD
AB=DC = 5—i^2 litres = 2XlO-3-m3
AD=BC = 2 — 1 = 1 atmosphere=l*0l3Xl05 Nt/m^
Work done = 2X 10-®x 113X10® joules,
= 202*6 joules
Expected Questions
1. la) Define the meaning of the terras (0 coefficient of increase of pressure
at constant volume {i7) coefficient of volume at constant pressure.
ib) Prove that in the case of a perfect gas the pressure coefficient is equal
to the volume coefficient. Explain how this concept leads to the
absolute zero of temperature. ^ ^
2. (a) Explain what is meant by a perfect or ideal gas. In what respects do
ordinary gases difier fiom a perfect gas ?
(6) State the various laws applicable to a perfect gas.
3 What is meant bv the characteristic gas equation ? Deduce the equation
PV=?nrT for a gram molecule of a gas from Boyle s and Charle s Laws. j ^ ^
4. (a) What are the units in which gas constants r or if are measured.
(6) What is the numerical value of gas constant in.
(i) Joules/kgm.— mole—
(it) ergs/gm.— mole— ®A'.
5 What is meant bv gm.-mole of a gas. What is its value lor oxygen.
is the volume occupied by one gm.-molecule of any gas at N.T P. Charle’s
6 (o) Deduce the equation. I assuming Boyles Lav, and Charles
Law.' What is the Physical significance of if.
i:> me i ^ aUa
Show that B represents the external work done by a nn.t ^ ,
cas when heated at constant pressure hroug •
7.
(a)
(b)
are their approximate values ?
CHAPTER IV
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
4.1. lotroductiOD. Heat is a form of energy and as such it is a
measurable quantity. So far we liave concerned ourselves
with the measurement of temperature and the changes produced
in volume b}' heat without any idea to measure the quantity of heat
required in a particular process. In this chapter, we will discuss the
measurement of heat and also the conditions upon which the transfer
of heat from one body to another depends. The process of measuring
the guaniities of heat is called calorimetry.
4.2. Units of Heat. To measure the quantity of heat the common
units used are (i) Calorie (cal) (ii) Kilo calorie (k. cal) (iii) British
thermal unit (B. Th. U.) (iv) Therm (v) Centigrade heat unit (C. H. U.)
Calorie. This is the unit of heat on the C. f?. S. system. It is the
quantity of heal required to raise the temperature of 1 gram of water
through I°C.
It may be pointed out that the quantity of heat required
to raise the temperature of i gram of water through is not the
same at all temperatures. For example, the heat required to raise the
temperature of i gm. of water from 10®C to 1 is not the same as
that required to heat it from 40®(7 to 41®C. For this reason if we take
one gm. of water and raise its temperature from 0*C to I00®C (at
standard pressure) and then divide this quantity of heat by 100. we
get what is called as the mean calorie. It has been found that the
value of this mean calorie is equal to the amount of heat required to raise
the temperature of one gm. of water from 14‘5°C to 15'5°C. For all
engineering purposes this unit is generally used. It is known as
15®caIorie.
Kilo-Calorie. This is the unit of heat in JJf.K.S. system. /( is
ihe amount of heat required to raise the temperature of 1 kilogram of water
through rc.
1 Kilo Calorie:= 1000 Calories,
73
74
ENGINEERING PHYSICS
4-3. British Thermal unit (B. Th. U). This is the unit of heat in
F .P .S. sj'stem. It is the amo^mt of heat required to raise the temperature
of one pound of trater through FF .
Since l lb of \vater=454 gms and \^F^— ®C.
9
1 B. Th. £7=454 X— ^ = 252 Calories.
Another bigger unit in the F.P.S. system is called Therm which
is equal to lO* B. Th. U.
1 Thermo 100.000,0 B.Th.U.
4.4. Centigrade Heat unit (C-H U.) It is the amount of heat required
to raise the temperature of 1 lb of ivater through 1°C.
1 C.H.U. = -^B.Th. U.= V%B. 252 = 454 Calories.
4.5. Specific heat. When a body is heated, it absorbs heat and
when it is cooled it loses heat. It is found that a cartain mass of a
substance requires a definite quantity of heat to raise its temperature
through a certain range. This quantity of heat is different for
different substances. Hence the quantity of heat ((?} necessary to
raise the temperature of the body
(/) is proportional to the mass of the body (m)
Qer.m
(ii) is proportional to the rise in temperature (f),
Q oc t
Combining both these factors, we have
Qozmt
where S is the constant of proportionality. Its value depends
UDon the nature of the substance but is independent of m and t.
This constant S is known as specific heat of the substance and is equal
to
9 .
mt
If m=l gm. and f=»I°CThen
Hence specific heat of a substance is equal to the quantity of
heat in calodes required by I gm. of that substance for heating i
through rC. Its unit is calories/gm/ C.
It can also be defined as the quantity of heat rneasured in kilo-
caloril for raising the temperature of 1 kg. of the substance through
1®C. In this case its unit is iC— cal/kg/ C.
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
The units chosen in F. P. S. systcrn are B. Tk. U.JlbVF and
C, H. U. fibre depending upon the scale of temperature ubcd.
Let m gm of a substance of specific heat .S\ require units of heat
when it is heated through (°C.
then Qi—7nSit •••('')
Ifmgm. of water of a specific *9, is heated through the same
range of temperature, then it will require units of heat. In this cast
Q^=wS,t
From (i) and (ii) we have
For water the value of specific heat is alway unity wViich is tlie
maximum value. Thus putting ^2 = 1 in the above equation, we
have
Hence Specific heat of a substance is the ratio of thr. quantity of
heat required to raisf the temperature of a certain mass of the substance
through a certain range of temperature to the amount of heat required to heat
the same mass of water through the same range of temperature. Thus speci-
fic heat
Quantity of heat required to raise the temperature of a given
mass of a substance through a certain range
~ Quantity of heat required to raise the temperature of the
same mass of water through the same range
Quantity of heat required to raise the temperature of unit
mass of a substa nce through
“Quantity of heat required to raise the temp^ature of unit
mass of w^ater through 1®C
The specific heat expressed as such is only a ratio and has no
units.
Specific heat of different substances is different. Specific heat of
water is maximum (unity) and so for the same mass and temperature
it has maximum energy. Out of liquids mercury has least specific
heat. In solid state specific heat is less than that of liquid of same
material. Specific heat of ice is much less than that of water. Specific
heat of substances generally increases with increase of temperature
(described later on). In an experiment mean specific heat is
found out.
a^nount
4.6. TJeimal capacity. The themuil capacity of a body is equal to the
int of heat required to raise the temperature of the body through rC.
ENGINEERING PHYSICS
If a body of mass m and specific heat S is heated through TC then
the amount of heat required by it is equal to its thermal capacity.
Thermal capacity=m5x 1 calories.
The units of the thermal capacity are cal/®C, k— cal/°(7, B. Th. 17/
and C. H, U.rC depending upon the unit of mass (whether gm. or
kgm or lb) and scale of temperature (whether \°C or chosen.
4.7. Water equivalent The water equivalent of a body is the amoiint
of water which absorbs or loses the same amount of heat as is done by the
■body when Us temperature is raised or lowered through 1°C.
Let m be the mass of the body and S its specific heat, then heat
required to raise the temperature through \°C is equal to mS calories.
This quantity of heat will raise the temperature of mS gms of water
through \°C, since specific heat of water is unity.
water equivalent =7 w 5 gms.
The units of water equivalent are gm, kgm. or lb.
Thus we see that the water-equivalent is numerically equal to its
thermal capacity but former (water equivalent ) is measured in gms.
while the latter (Thermal Capacity) is measured in Calories.
Quantity of heat required to heat a body- Let S be the specific
heat of a given material. Then the amount of heat required to raise
the temperature of m gms. of that material through FC is given by
Q=mSt.
The same relation will also give the quantity of heat lost by a
body when it is allowed to cool through a certain range of tempera-
ture. Then Q will be the amount of heat lost by the body and t,
the fall in temperature.
4.8. Principle of measurement of heat. ° ^\he
different temperatures are mixed together, heat flows
substance at a higher temperature to the substance at a lower
temperature till the temperatures of the two become the same. I
S ?Socessno heat has been gained from or given to any
oXide body and there is no chemical action in the mixure, then fAe
heat Ll by tL hot body must be equal to the heat gamed by the cold body.
Thus
Heat Iost=Heat gained.
The above equation is termed as the principle of calorimetry.
TT 1 ^ boiler in a heating system is 50%
Example. • f ^ After burning th'i coal which liberates
boiler is filled with water at ’ > f 4..^^ 210°F What is the
50,000 B.Th. U. the water has a temperature of ^10 ^ ^
weight of water in the boiler .
Quantity of heat «=» S t where S is the specific heat of water
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
For water S— 1
Q=mt.
Note [Q is in B. Tk. XJ. if m is in lbs and ( in
Total heat produced=50,000 B. Th. U.
Heat utilised = x 50,000 = 25000 B. Th. U.
25000= m (210 — 65)=mX 145,
25000
145
172*4 lbs.
Example. 2 . In two experiments, 610 gm. of wafer are heated from
14*^C to and 30 lb 8 oz of water are heated from o4^F to 89^F.
In which case is the greater quantity of heat supplied.
[A. M. I. E.)
(*) Mass of water heated =610 gms.
Rise in temp «80*C.
Heat supplied Hi =m 8 610 x 1 X 80=48800 cals.
(u) Mass of water heated =3*5 lbs.
Rise in temperature =55®F’.
.*. Heat supplied Ha 5 f =3*5 X 1 X 55 = 192*5 B. TA.
But 1 B. Th. V. =252 calories.
Hg =192.5X252 =48510 calories.
Hence. Hi>Hg, i.e. more heat is supplied in the first case.
4.9. Methods of measoriog specific heat of solids.
The following methods are available for the determination of
the specific heat of solids.
1. Method of Mixtures.
2 . Method of Fusion of ice (Bunsen’s Ice Caloriemeter)
3. Joly's Steam Calorimeter.
4. Nernest and Lindemann’s Vacuum Calorimeter.
1. Metttods of mixtures. The specific heat of a solid such as lead
shots or copper can be easily determined by a simple method
detS?nf method of imxtures. The soUd whose specific h^at is to bt
deteimined is placed m the copper tube of the hypsometer and mouth
of the tube IS closed with a cork having a hole . Through the hole a
thermometer is passed so that its bulb and a part of its stem are
»..K rsSi: s
i
7S
ENGINEERING PHYSICS
iilled nearly one-third with cold water having a temperature about
10®C below the room temperature and weighed again. The tem-
SCREEN
\J
N
N
s
s
s
s
S.
N
V
s
S
s
N
\
s
s
s
Fig. 4.1.
■perature of heated solid in the hypsometer is noted when it becomes
i^onstant The temperature of the cold water in the calorimeter is now
noted and the hot solid is quickly transferred into the calorimeter.
The mixture is well-stirred and its final maxmium temperature is
recorded. The calorimeter is once again weighed with its contents
With the help of the above observations, the specific heat of the
solid can be calculated as follows.
Let the mass of the calorimeter and stirrer =m.i gms.
Mass of the cold water taken in the calorimeter =m gms.
Mass of hot solid added.
Initial temperature of cold water and calorimeter=^ C
Steady temperature of hot solid — ^ ^
Final temperature of the mixture — ^
Let S be the specific heat of the solid and s that of t e ma er
of the calorimeter.
Heat lost by hot solid in cooling from T C to 6
•SPECIFIC HEAT OF SOLIDS AND LIQUIDS
79
Heat gained by the cold water, calorimeter and stirrer in raising
the temperature from t°C to 0®C*.
=m [6 — {d — t)
= (m+ rriys) — /)
Since Heat lost =Hcat gained
.*• MS {T — 6) =(?rt+mi5) (6 — i)
or specific heat of tne solid
_ (w+wis)(g— /)
M [T—d)
Example. 3- Apiece of iron of mass 100 gms is rapidly ixmovid
from a furance a'nd immersed in a calorimeter weighing 46 gms. The
furnace temperature was and the calorimeter contained 86’5 gms
of water at The final temperature of the contents of the caloriemeter
was found to he 22°C. Calculate the specific heat of iron. The specific
heat of the calorimeter is O’l.
[A. M. I. E.)
Mass of iron lOO gms.
Temperature of iron T= 98-5®C'.
Mass of calorimeter,
Mass of water
46 gms-
w= 85*5 gms.
Temperature of water t—l 5^0.
Final temperature of the mixture O^il^C
Specific heat of the calorimeter = 0’1
Let S be specific heat of iron.
Fall in temperature of iron =(r — 6) ==98-5 — 22=76*5°C.
Rise in temperature of calorimeter and water
={e—t) =22-~\5=ro
Heat lost by iron ^MS {T—d) =100x5x76.5 cals. ...(i)
Heat gained by calorimeter s (0— /) = 46 X 0* 1 X 7 = 32 2 cals.
Heat gained by water = m {O-t) =85-5 x 7 = 598-5 cals.
Total heat gainet by calorimeter and water =(-mi 5 + »») {8 — /)
= 32*2 +598-5 =630*7 cals
But Heat lost=Heat gained.
I0OX5X76-5 = 63O'7
or
630*7
7650
= 0-0825
.nf calorimeter weighs 200 gm. and contains 1000 am
“ f of 22»0. If 400 gm. of a mixture of copper and
ZtZ “ <e«pero(«re of 100 “C ore put into the calZmeter
80
ENGINEERING PHYSICS:
[Specific heat of Al = 0*218.]
Mass of the calorimeter
mi = 200 gms.
m— 1000 gms.
<= li^C
9=
s~ 0*093.
mass of water
Temperature of water
Final temperature of the mixture
Specific heat of the calorimeter
Wt of the mixture of copper and aluminium
M •= 400 gms.
Specific heat of 5'=0.218.
Heat absorbed by water and calorimeter «=(m + rTij s) — <)
= (1000 + 200 X 0.093V
(26*5— 22)-
= 1018*6X4*5 = 4583*7 cals
[Note Calorimeter is always of copper material hence specific
heat of copper s~ 0*093.]
Let w be the mass of aluminium filings in the mixture so that
(400— is the mass of the copper filings.
Heat lost by Aluminium=uj5(7 ’ — ^)=ujX0'218x (100 — 26*5)
= w XO-218 X 73*5 cals
Heat lost by copper = (400 — w) s x { T 9)
Heat lost by copper = (400 — w) 0*093 (73*5)
Total heat lost by mixture =«;X0-218 X 73.5 +(400— w^)
of copper and Al. filings 0* ^93 X 73*5
_}_ 2809 — 6*85 m'
*=9*25uJ + 2809.
Heat lost =Heat gained
9-25 1C + 2809 « 4583*7
or u; —
1774.7
9.25
--192 gms
2 Bunsen’s Ice Calorimeter. It is a method for determining the
specific heat of a solid by using the phenomenon of fusion of ice.
The fact that ice conlracta on melting was utilised 1^ Bunsen m
consSttinf I very delicate -lorime^^ kno^ as ^1-
Calorimeter. It has ^^J^ater at 0 ®C^s a volume of 1*00001 c.c.
1 0908 c.c. whereas 1 ^* decreases in volume
Hence the fact that one gm. of ice on ^.o^rect weight of
by (1 0908 — 1*0001)=0*0907 c.c. is used to find the correct weignt u
ice melted by the hot solid.
part of w'liich a
Fig. 4.2.
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
It consists of a wide glass tub: B into the upper
long thin walled test-tube A is fused.
The lower end of the wide tube B is
sealed at C to a narrow tube CD bent
twice at right angles. The other end D
of this tube terminates into a collar
which is closed by a cork througli wliicli
passes a bent piece of capillary tube of
uniform bore. A scale is attached to
the horizontal part of this capillary tube
so that volume between any two divi-
sions is known accurately. The upper
part of the wide tube B is filled with
pure water from which dissolved air has
been removed by boiling. The remain-
ing part of the wide tube 5 and whole
of the bent tube CD and also a part of
the capillary tube are filled witli pure dr>' mercury
irp now placed in a vessel containing finely broken
attain the temperature of ice. Some^ previo-
intn it ^ ponred into the tube ^ and air is^ blown
is removed and ice cold wat^er is introduce?!^^
. .,„d, ,0.1, id. ,,i,ioh‘ r;;.;’o’„ (ilo 5,rr'’'
specific heat io to be determined is t;,L-
and weighed. It is heated to a constant high temp^ratie Jh
then suddenly dropped into the water in A and the tube is mrk .h t
once.The heat contained in the solid body will be given to the snrr
tS! ‘"n contained in the tube Tnd win me p rt of Tt'
This will continue till the temperature of the solid and water in Vh
tube IS again o''G. With the melting of ice contraction in thn
^es place which causes the mercury meniscus in ^he capmar^ tn h"
5.”. pT.?,io;"r;t,»a “» -i ‘•s.-r/*"
decreases ^rvolume t; 0.0907 cc thrma-^'^ftf "'‘='‘‘"5
decrease i„ volume equa^ to Jl is given “
SI
V
0-0097
H L cal/gm. is the latent heat of fusion of ice then heat gained
S2
ENGINEERING PHYSICS
by m gms. of ice at 0 ®C in melting=/nZ/
L’V
"^^907
If M is the mass of the body, 5 its specific heat and t its initial
temperature, then
Heat lost by the solid in cooling from f’C to 0 °C=MSt ca!s.
As this has fully been utilised in melting ice, we have
Lv
MSt=
5 =
0*907
Lr
0-0907
Advantages.
1. It is very sensitive and accurate method.
*> This method is specially useful for solids which are available
in small quantities.
3. The water equivalent of the apparatus does not enter into the
calculations.
4. There is no radiation taking place as the whole apparatus is
surrounded by ice which is a poor conductor of heat.
5. The apparatus can be used both for solids and liquids.
FxamDle 5. If one gram of ice at 0°C contracts by 0 091 c.c. on
melting calculate the mass of a metal of specific heat 0'112 heated to
lOO^C u'hich when dropped into an ice calorimeter causes a decrease m
volu^fic of 0'0G3/
Decrease in volume of ice 1^2=0-0637 c.c.
Decrease in volume when 1 c.c. of ice melts
= 0*091 c.c.
Let m be the mass of ice melted.
w =
0-091
0-0637
0-091
Temperature of the metal. (= 100®C
Specific heat of the metal, >S'=0-I12
Let J/ be the mass of the metal dropped into the ice caloiimcter.
Heat lost by the metal ^MSt
=d/x 0-112X100 cals.
Heat gained by ice
Heat lost
J/XO-112X100
ft ft
or
£=mL= 7 rtX 80 cals.
=Heat gained
= 0-7 X 80
0-7 X 80
0-1 12 X 100
= 5 gms.
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
83
Example 6. The density of ice is 0‘93 gtn. per c.c. at Q^C. A piece of
metal weighing 150 gms. is heated to lOO'^C and is then placed in Buyiseti's
ice calorimeter. The decrease in volume is found to be 1'876 c.c.
Calculate the specific heat of the metal, latent heat of water being 80.
Density of ice at 0®C —0*93 gm./c.c.
Volume of i gm. of ice == — V075 c.c.
WTien this l gm. of ice melts, it forms l gm. of water wliose
volume is l c.c. In other words, when 1 gm. of ice at 0®C melts to
water at o^C, the contraction in volume is
I'U/a — 1 =r
• Contraction in volume as given =
/. mass of ice melted =
u u / 3 c.c,
1*876 c.c,
1*876
0-075
= 24*924 gms.
Heat gained by ice for melting =^mL
= 24*924X80 cals.
Hass of the metal piece il/=I50 gms.
Temperature of the metal piece t=l 00®(7
Let S be the specific heat of the metal piece.
Heat lost by the metal =MSt
= 150x5x 100 cals.
Heat lost=Heat gained
150 X 5 X 100 = 24*924 X 80
^_ 24*924 X 80
150 X 100
=0-1329
Example 7. A substance weighiny 20 gtns and nt lOfi°n
dropped inia a Bunsen ice colorimeter and /I. r j K C* tww
«ipillary tube of 1 SQ. m.m in area column in the
-siJcct^c heat of the suhstance. ’ through 5 cm. Find the
Owen that 1 gm. of ice on melting coMracta hy 0-09 c c
Mass of the substance if—
Temperature of the substance , oo»^*’
Contraction in length of the column of liquid I
Contraction in volume
^axl
, =0*01X5=0*05 C.C.
ENGINEERING PHYSICS
Mass of ice melted =
V
0*05 5
^ « • cm
i\*nc\ rt O
Heat gained by ice
0093 009
=0*555 gm.
—m X L
= 0*555 X80
=44*4 cals.
Heat lost by the substance
=3/67 = 20 x 5 x 100 cals,
Heat lost = Heat gained
2000 5=44*4
Example 8. ‘-^0 gms. of water at lo^C are put into the tvhe of
Bunsen's ice calorimeter and it is observed that the mercury thrt ad moves
through 30 cm. 12 gm of a metal at 100°C are then placed in-
the tube and the mere ury thread moves through 12 cm. Find the specific
heat of the metal.. /rr d v
(C/.i ./>•)
(0
{ii)
Mass of water
Specific heat of water
Initial temperature of water
Final temperature of water
Mass of metal
Initial temperature of the metal= 100 ®C.
Final temperature of the metal =0®C.
= 20 gms.
= I
= 15°C.
= 0°C.
= 12 gms.
{i)
Specific heat of the metal
Heat lost by water
=5.
=Mass of water X Specific
heat X rise in temp.
= 20X I X 15 = 300 Cals.
= 12 x 5 x 100 Cals.
= 1200 5 Cals.
The movement of mercury thread for loss of 300 Calories=30.cm.
The movement of mercury thread for loss of 1200 S Cals — 12 cm.
300 12005
(u) Heat lost by metal
30
12
or
X
12
1200
-= 0*1
rSPECJFIC HEAT OF SOL'DS AND LIQUIDS
«5
Example 9. Jf the latent heat of fusion of ice is 80 , a^id its density
at O^C is 0 017 , find the travel of the mercury in the tube of a Bunsen ice
■Calorimeter, when 10 calories, are given to the ice, the diameter
of the tube heimj, 0'4 cm.
{Land. IJniv.)
Quantity of heat supplied to ice 9== 10 cals.
-Amount of ice melted
Q
^gms.
Density of ice at O^C P = 0'917 gms/c.c.
Volume of ice melted Fi = —
^ P
Now 1 gm. of water has a volume of i*000 c.c.
Volume of water formed V=m cm^.
C'-i. ^^iminution in volume due to melting=r,— F
aUtance which the
Change in volume of the mercury in the tube=rrr^f.
Hence
fl— Pi — I0X(1— 0*917)
TT X 4 X 1 0-* V «A n.o . =9 00 cm.
X 4 X 10“* X 80 X 0*9 1 7
■an interned diametePof^T^!^ WUn a*^i2ef calxirimeter has
mat is the specific keat % the Jttll nwves 4 cm.
./ z,s zfo!
• gm of^water has a volume of f.ooo c.c. & , gm. of ice has a volume
of
1.9,7 c-c. or 1.091 C.C. Thus a contraction of 0-091 c. c.
AM « *
occurs when i •
■=rr ( 0 - 02)2 x4=0-ISl6m^ ^he contraction=w 2 f
Mass of ice melted
m
_ 0*00I6:r
gm.
0*091
86
ENGINEERING PHYSICS.
Heat gained by nietai, mL =
0 0016ti:X 80
0*091
Heat lost bv ice =3ISt
Heat lost =Heat gained.
O.OOI6-X 80
0.091
= 0-5 XSX 100
or
0*0016t:X 80
0^91 ^*5 X 100
= 0 088 cal/gm-
3. Joly’s Steam Calorimeter fhe principle involved in this method
is tliat tlic heat necessary to raise a body from the room temperature-
to that of steam is measured by the mass of steam condensed into-
water at the same temperature to supply that heat.
JOLY STBAM CH£miMET£P
Fig.4.3.
The app^atus
^^^?^"?inl^and^s^pr?vidld wfth suitable holes for letting steam in and
rs',., s,
;,'A’„'ch"X”lr“mVn. .™ ol . a.M. b.M« by
order to prevent “X "uspension wire. Formation
drop fn HTiZLr prevented by electrically heating the
SPECIFrc HEAT OF SOLIDS AND UQblDS
87
suspension wire to the temperature of steam with the help of
heating coil P immediately above the hole H. There is a shelter
G provided at the top of the chamber so that steam condensed on the
ceiling of the chamber may not fall on the pan P.
The mass of the substance when there is only air in the chamber is
determined. The substance is tlien allowed to remain in the pan for
some time so as to take up the temperature of the chamber and its
temperature is noted with the help of thermometer T. Steam at full
pressure is suddenly admitted into the chamber from tube .4. which
after circulating through the chamber escapes through the outlet B at
the bottom. A rapid flow of steam through the chamber is necessary,
essentially at the beginning, in order to prevent partial condensation
of the steam due to tlte radiation to the cold air and walls of the
chamber. Steam starts condensing on the pan P and weights arc added
on the other pan to maintain the balance, .-^fter five minutes or so
when P ceases to increase in weight the temperature of the steam is
noted. Now finally steam supply is reduced in order to avoid steam
currents which would interfere with the accurate weighing. The differ-
ence between the two weights used for counterpoising at the start
and end of the experiment gives the weiglit of the steam condensed
on the pan P.
Let m — mass of solid substance in gms.
3/=maRs of the steam condensed on the substance and the
pan in gms.
fi=initial temp, of solid substance i.c,, of steam chamber.
f 2 =temperature of steam.
w = water equivalent of the pan etc., in gms,
S =specific heat of solid.
^=la^tent heat of steam in calories per gm.
Heat given out by steam during condensation on the substance
and the pan=j»/L Cals
Heat gamed by the substance and the pan = (m5+*r)(/2 -/i) •■.(»)
Heat Lost = Heat gained
Thus, we have
Hence specific heat S can be found out.
engineering physics
V small mass
method ^ substances can be determined by this
(2) Since the final temperature of the body is the same as that
of Its surroundings, hence no “cooling correction’* is to be applied.
f3) The only reading required in the experiment is weighing which
can be done rather very accurately.
(4) It is equally applicable to powders, liquids and gases. In the
case of , powders and liquids, glass or metal containers are used
and thermal capacity is taken into consideration. For gases a modified
and improved form of the apparatus is used as will be explained
in the chapter on specific heat of gases.
4. Nemst and Lindemann’s Vacuum Calorimeter Method
Nernst and Lindemann have used the electrical method for
measinng the specific heat of solids at low temperatures. Two types of
calorimeters are employed (i) for good thermal conductors (n) for bad
thermal conductors. For good conducting solids, the calorimeter is made
of the solid whose specific heat is to be
determined. The cylindrical block of
the metal .4 is drilled, and a cylinderi-
cal plug B of the same metal is fitted
in it. B is shaped at the top so that
good thermal contact with A is established
The platinum coil H is wound on para-
ffin waxed paper wrapped round the plug
B and is kept electrically insulated
from A by filling the inter space with
paraffin wax. The platinum coil serves
dual purpose.
(t) as electric heater to produce
heat and
l-ig. 4.4.
(u) as platinum resistance thermo-
meter for measuring the initial and final
temperatures of the metal A.
The whole arrangement is then
suspended by the connecting leads L and
L' in glass flask D. inside a Dewar flask F
filled with ice or liquid air or liquid hydrogen according to the low
* ^o+iirp renuired Hydrogen which is good conductor of heat is
fi^^I^admitted into D and the metal eventually assumes the same
first admitted 'nto^ an^^^. ^ completely
er^Ite" and sealed, so that loss of heat by conduction or
convection is almost entirely eliminated,
89
:SPECIFIC HEAT OF SOLIDS AND LIQUIDS
To determine the specific heat at a given low temperature, a
current / at a known potential difference V is passed through a
platinum coil for i seconds.
Let and 7, be the initial and final values of current in amperes, th n
V T'
and i? 2 = 7 “ initial and final resistances of the
platinum coil are readily determined. Then, applying the relation
(1 -i-oidd) where a is the temperature coefficient of resistance of
■the coil, the small rise of temperature <16 of the wire which is usually
1®C can be found out. From the average value 7 of the current the heat
produced electrically is
•equivalent of heat.
Vlt
calories, where J is Joule's mechanical
Let m be the mass of the calorimeter i. e., cylinders B and A and
.S their specific heat, then
Heat produced electrically = Heat absorbed by the solids (8 & A)
-f-Heat lost by radiation.
Vlt
=9nSd6~\-h
Neglecting the small loss of heat due to radiation, we have
S =
Vlt
Jmdd
are bad conductors of heat, the calorimeter some-
What modified is shown in Fig 4.5. It consists of a hollow cylindrical
silver vessel d on which the coil H of platinum wire is wound The coil
X°vessel 'v “h'isTh is plLe'^S
air inside d from escaoW Ti^l r.r! the
vessel is essential for rat>idlv inside the silver
equilibrium inside the vessel Th^i conditions of thermal
the leads i i of platinum ioiM^ 'a ‘ n suspended by
90
engineerjng physics
Advantages, (j)
Y’O PUMP
Since whole of the calorimeter is enclosed in
vacuum the exchange of ehat with the
surroundings is very small.
(2) Since the rise in temperature {d$)
IS very small, this method gives .the
specific heat at a particular temperature
instead of the mean value over a certain
range.
It may be noted that the second type
of calorimeter meant for bad concuctors can
be used for the determination of the
specific heat of liquids and gases.
The apparatus though has been
designed for low temperatures, can also
be used at ordinary temperatures.
4-10. Specific Heat of Liquids- The
specific heat of liquids may be found by
tlie following methods
(i) Method of mixtures-
(ii) JouIe^s Electrical method.
(iii) Callender and Barnes continuous flow method.
(iv) Method of cooling.
1. Method of Mixtures- In this case the procedure is exactly the
same as in in Art. 4.9 except that a solid of a known specific heat is
heated to a constant temperature and transferred to the calorimeter
containing the liquid whose specific heat is to be determined*
If s is the specific heat of liquid, then
Heat gained by the liquid and calorimeter.
= (w + ms) { B — f)
and Heat lost by the solid = MS {T — 0 )
{w ms) {B — 0 “ S [T B)
or Specific heat of liquid.
{T — B) w
m {B — t) m
\
Where JI = mass of the hot solid
S = Specific heat of solid.
_ The temperature of the hot solid.
t = Initial temperature of the liquid.
8 = Final temperature of the mixture
w = water equivalent of the calorimeter =
where = mass of the calorimeter.
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
Example ll- 200 gm. of water a' were mixed vnth 200 c.c. of
milk of density 1‘0Z at 50*^0 contained in a brass vessel of thermal cajMCity
equal to that of S gm. of water and the temperature of the mixture was
64'^C. Find the specific heat of milk.
Mass of water =200 gm.
Fall in temperature = (98 — 64) = 34®C
Heat lost by water = 200X 1 X 34 6800 cals.
Mass of milk =200 X 1-03= 206 gms.
Rise in temperature = (64 — 30) = 34°C.
Heat gained by milk = 206 x X 34
where S is the sepcific heat of milk.
Water equivalent of brass vessel = 8 gm.
Heat gained by brass vessel = 8 x 34 =» 272 cals.
Total Heat gained =(206 xSx 34 + 272) cals.
Heat lost = Heat gained
6800 = (206 X S X 34 -h 272)
5 =
6800—272
206 X 34
= 0*932 cal/gm
2. Joule’s Electrical Method. This is a laboratory method and
employs electrical heating. The apparatus consists cf a heating coil
H suspended in a copper calorimeter covered with an ebnoite disc.
The ends of the coil are soldered to copper leads which are further
connected to two binding screws Si and S.^ fixed on the top of the lid
as shown. The lid has two holes, through one passes a sensitive
thermometer and through the other a stirrer. The calorimeter is placed
in an outer jacket which is packed with cotton wool to minimise loss
of heat by conduction and radiation. The terminlas and S 2 are
connected to a rheostat R, ammeter A , battery B and key K in
series. A voltmeter V is connected in parallel with the terminals Si and
S^ as shown.
is
First of all the weight of the calorimeter and the stirrer with lid
taken. Now, the calorimeter is filled r
nearly half with the liquid whose specific
heat is required. It is weighed again to
find out the weight of the liquid. The
lid is placed with the coil H in position
so that the coil is immersed in the liquid
without touching the sides of the
calorimeter. The liquid is kept stirred by
a suitable stirrer.
After closing the circuit, current I
and potential difference V are adjusted in
such a way that rise in temperature is
about in one minute. Key is taken
out and after stirring the liquid well,
its initial temperature is noted very
accurately. The key is inserted and a
constant current of I amperes is made to
Fig. 4, 6,
k
engineering physics
pass tlirough the coil for the time the rise of temperature of
tlie liquid of about lO^Cis obtained. The readings of ammeter
and \oltmeter are noted. The time in which this rise of temperature
lias occurred is noted with the help of stop watch.
Let m=mass of the liquid
<S’— specific heat of the liquid
w=water equivalent of the calorimeter
?i=initial temperature of the liquid
^2 = final temperature of the liquid
Tr^potential difference across the coil (reading of volt-
meter T)
/^current through the heating coil (reading of
ammeter)
/=the time for which current passes.
Heat produced electrically= ^ calories.
whereV is the mechanical equivalent of heat =4*2 joules/cal.
Heat gained by the liquid and calorimeter=(?6’+m5)(^2“^i)
If h is the heat loss from the calorimeter, then
If the rise of temperature is small, the loss of heat due to
Tadiation is negligible. Hence neglecting the loss of heat due to
xadiation, we have,
fi)
Vlt w
or S= n"
mJt m
where (rise of temperature)
3. Callender and Barne s Continuous flow method. It is based on the
same principle as joule’s method. The apparatus consists of a heating
coil of fine platinum wire H in the form of a spiral. The spiral fo^
of coil helps in two ways (t) It increases surface area exposed to the
liquid, (ii) It helps in thorough stirring of liquid as it flows. The
platinum resistance wire is mounted in a narrow glass or quartz tube
T (2 m.m. in diameter) with its ends fixed to t\vo thick copper tu^bes
The liquid whose specific heat is required is made to A^w
continuously through the tube T at a steady rate and is heated by
passing a current through platinum resistance wire by means of
battery circuit connected to the copper tubes as shown in Fig. 4.7.
In each copper tube is inserted a bulb of platinum ;«istance
thermometers Pi, P 2 to measure the temperature of the ^
entering and on leaving the tube. The copper tubes serve two
SfSaFIC HEAT OF SOLIDS AND LIQUIDS 93-
important purposes (i) due to their high conductivity, they
keep the whole bulb at the temperature of the surrounding
LIQUID T T WATER i LiaUiD
Fig. 4.7
liquid and (u) due to their low resistance, the}* do not
let an appreciable amount of heat to be produced near the bulbs of
the thermometres. The tube T has two openings, one for allowing
the liquid to flow into the tube and other for allowing it to go out of
it after being heated. This tube is enclosed in a vacuum jacket V
which is further surrounded by another jacket through which water is
continuously circulated. This arrangement is made to minimise loss of
heat due to conduction and radiation. The tube is closed on botli sides
by non-conducting lids E and F. To corresponding ends of both the
copper tubes are connected battery B, rheostat B. ammeter A and
key K. The voltmeter is connected in parallel with the spiral.
The liquid whose specific heat is to be determined is mad^
flow through the tube T at a constant rate. The kev A' U
inserted so as to close the circuit. The electric current now flows
through the platinum spiral & thus the liquid is heated on its i>ass i^e
through the tube. Sufficient time is allowed for the aonar^+n-f
settle down to a steady state in which case all the heaf nmdnr^H
electrically is carried away by the liquid and none is used up bt the
apparatus to raise the temperature of any of its parts. When a sfeaHv
state IS reached as indicated by the constancy in the readings of thl
Let »(=mass of the liquid flowing in t seconds
t - --licated by
jS=Specific heat of the liquid
F==Voltmeter reading in volts
/=Ammeter reading in amperes
Heat gained by liquid^mS
Heat produced by current in time t seconds
engineering physics
f>l
Vlt
calories
If h is the loss of heat due to radiation, then
(#2—
To eliminate the radiation loss, the experiment is repeated with
voltage T', current /' and a different mass m* oi the liquid for the
same time t, over the same temperature range (? 2 — t^^n
l^=m:s (“)
Subtracting (i) from (n) we have
J =: (m — 7rt) 6 (t'2 — ei)
or
{V' V-Vl)i
J {m'-m)
^ *••/> m X/
Advantages («) Since the flow is steady there is no change of
temperature in any part of the apparatus during the experiment and
hence no correction is required for the thermal capacity of the
calorimeter.
ItA The temperature of the inflowing liquid can be measured
more' accurately by using platinum resistance thermometers which
would have been otherwise useless if changing temperatures were to
be measured.
{Hi) Radiation losses are almost eliminated.
Rv choosing suitable values of V and J. the rise in tempe-
ratur': ^can "be Zr small so that specific heat of liquid at any
temperature can be found out. •
(v) This method can be used for
Vipnt of water with temperature. Callender and Barne s also
this method for finding the^ecific heat of mercury, the spiral coil
Tn this casTwafd^^^^^^
^"";")"Th!s“method' is used for the determination of Joule’s
mechanical epuivalent of heat.
Fxamole 12. It is observed that the temperature of 300
U^ui^Tltunm flas^ of -'"f oTTotZ^i^^e
J = 4-18 joules! cal.
Mass of the liquid «»=3<)0 gms.
Water equivalent of the flask «« 2 gms.
Rise in temp. " rZJo otais.
Resistance of heating ^ ^
Current passing
95
SPECIFIC HEAT CF SOLIDS AND LIQUIDS
Time
Specific heat
But
Substituting the values, we have
(l-8)=^X 10X3600
~ 300X 180X4*18
= 0*517— 0 067 = 0*45
l=lhour=3600 seconds.
^ Vlt w
~ mJ{t2 — ti) m
V-^IR
5=
I-Rt
mJ{t2 — li)
w
m
20
3^
Example 13. In a continuous flotv experiment, liquid passing at a
rate of 10 gmimt. over heating ,coil is heated from 22°C to 32‘^C when the
potential difference between the ends of thi coil is S volts and current U
I'OA, when the rate of flow is changed to 4 gmj^nt. the same rise of
« potential difference of 1-5 volts and a current
Calculate {i) the specific heat of the liquid and (ii) the rate at which
neat is lost from the surface of the tube carrying the liquid. Take J:=4-7f<
joules leal.
For the first observation.
Vlt ,
where k is heat loss due to radiation
For the second observation.
VTt ,
—j— ={it;+m's) (<2—
From (i) and (ti) we have
„ {V'r~Vl)t ^
''’here ^=/ 2 -«i=rise of temp.
_ (3X 1*5— 1*5X 1*4)1
4 T 8 (10/60 — 4/60) X Cal/gm/®C.
Substituting this value of S' in equation (i) we have
3X1*5 , 10
" 4 ^= (Oi-^XO-575) (I0)+A.
or h=i
3 X 1*5
iVSJ
- X 0*5 75 — 0*125 cal/sec.
4*18 60
and current flowing fmjg amus Thp Jf. ^ , the wire was 2-5 volts
is 5^C when the fate of flZ is 32 L? of the liquid
then increased to 50 gms. per mt the nnfpnr^
the current to 2*5 am«r to 3 volts and
before. Calculate the specific heat of the
96
ENGINEERING PHYS CS
First Case.
Temperaluro difference — ^“5’C
Mass of liquid flowing per mt. m=l2 gms.
Current flowing
Potential difference
2nd. Case.
Mass ol liquid flowing per int.
Current flowing
Potential difference
/=2'0 amp.
l'=2’5 volts.
m' = 50 gms.
r=-2'5 gms.
r'=.3*0 volts.
As the rise of temperaluro in the two cases is the same,
[rn — no')J[tf 2 —
(3 X2-5— 2-5X2) 60 ^
(50-32)x5x4-2 6 ' '
4 - Method of cooling. This method makes use of Newton s lau
of cooling. \Ve shall therefore, flrst make a detailed study of this law.
Newton's Law of cooling. When a hot cup of tea is placed for
sometime it becomes cold because it loses heat and air suirounc-
ing it gains lieat. .\t first it loses heat quickly but later on the loss ol
heat becomes slow. Newton put forward a law regarding the rate ol
loss ot heat and it is called Newt( n’s Law of cooling.
Newton s Law of cooling. It staUcs that the rate at which a body
loiies fuMt directly proportional to the temperature
the body and that of the surroundings. The amount of heat radiated
depends upon the extent and nature of the radiating surface. ,
Let 6 be the temperature of the body at a^y
l)„ tlie tempi-rature of tlio i, proportional to tlie
Newton’s Law of cooling, heat f, yf f^fl of
temperature difference («-»„). It follow^ that the rate
temperature will also be proportional to
Hence — ^
at
(IQ U /D
»o)
Where .IQ is tl;e amount of heat lost in time dt and k is a
ronstmrt depending upon the area and nature of the surtace ihe
negative' sign shows that the. temperature falls as time lapse .
The experiments have revealed that if the body ^
I rit air the law holds good for small differences of tempera
£i“e^c?:o"n.
" 'IL is the mass and S the specific heat and de is the fall of
temperature in the time dt, then
dQ=in 8 dd
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
(le
de k
97
or
dt “
where K is another constant and is equal to
dd
Jl
mS
or
or
d~e
0
= ~~Kdl
r dd r
or Jog {d—e^) = -Kt-\-C.
Where C is a constant of integration. This equation is of tl.e
VE5SEL-
time
Fig. 4.8.
‘rr "
expeS" ^ oan b’f by the to.Iowing
nearly Jdh wTth the hot outside is taken and is filled up to
“^4% ' Thil"^P^"t‘* ■" " ^bT-Cri vesTel a^'sh
iti rig. 4.9. iiiis vessel containc j t shown
immersed in it to find temperat^e of tU ^ thermometer
is noted after an^iite'^rTal oVra'^^rnuf ?mfh‘° T”'' t'^^tr^ture
approaches the room temperatme The^« *^"^Perature of water
found at different intervals of a "^^^^^ence of temperature il
» i. 5,. “f"”* “ by pi„i.s
98
ENGINEERING PHYSICS
{iii) If a body cools in still air by convection only, then this law
is true upto 30°C difference of temperature.
Specific heat of liquid by the method of cooling. If a liquid, heated to
a temperature above its surroundings, is put m a calorimeter and
allowed to cool in an enclosure, then the rate of cooling depends upon
the following factors.
(i) the temperature difference between the body and the
surroundings.
(u) the area of the radiating surface,
(iu) the nature of the radiating surface.
It should be noted that the rate of cooling does not depend upm
the nature of the liquid. Thus if the above factors
the same for hco different Uquids, their rates of^ohng W be
This is the principle of the cooling method employed for the
determination of specific heat of a liquid as given below.
Two exactly similar, small copper calorimeters blackened from
Fig. 4.10.
Fig. 4.1 1.
outside and having the same therntal capaciti« ^^are takem^ A
double walled vessel with water e , when empty are
constant temperature enclosure. The enclosure as shown
weighed and are afterwards insidyhe enclosure
in Fig." 4.10. Now water and given I'T"" to 35'C
be determined are heated to a temperat jj,
the calorimeters. The calorimeters “d tte ^„„eters T,.
by a lid provided with small holes to ?" water and the
and y, and stirrers inside the temperature of each
liquid are kept stirred continuously and th t Pe of the
after every half minute interval is recordea. a the mass
observations, the two calorimeters are r g j. ^ te found out.
of water and that of the =ame volume of the liqmd
The cooling curves are then plotted between t t t,y the
perature in each case as shown in Fig- 4 li.
99
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
through a certain range of temperature
times^beV^n^ / [■ are found from the curves. Let these
limes be and respectively and
wti=mass of liquid
wi2=mass of water
MJ=water equivalent of each calorimeter
‘S'=specific heat of liquid
Total heat lost by the liquid and the calorimeter
=m.^f (e^~e,)+w
^?ate of loss of heat of the liquid
h
Total heat lost by water and the calorimeter
=*n2 Xlx(e,- 92 ) +«,(»,_ 9^)
=(»i2+u.)
Rate of loss of heat of water
= K+w)( 9 ,- 9 j)
t
2
Since the rates of cooling are equal we have
^ (w3+w) (e,~~e,)
t
or
or
w)={?n2-j-Ti’) —
/a
g XV
fn.
(*)
• • •
(•■•■)
b. I "" "> <b.
I.";" ”« b.
f di;'””" ii't fh"
100
ENGINEERING PHYSICS
Example. 15. '250 (jms. of water and a7i equal vohune of alcohol of
mass 200 gins, are placed successively in the same calorimeter and cooled
from OO^C to 5-yC in 130 seconds and 67 seconds respectively. Find
the specific heat of alcohol. The water equivalent of calorimeter is
10 gms.
Water equivalent oi calorimeter iv
Mass of water ^”2
Mass of same volume of alcohol «ii
Fall in temperature (^ 2 — ^ 1 )
Time taken by water to fall through
5^C, h
Time taken by alcohol to fall through
Let the specific heat of alcohol be 6, then
= 10 gms.
= 250 gms.
= 200 gms.
= (60-55} = 5®C.
= 130 seconds,
=67 seconds.
«’
■m 1 1
1'2
■m
(250+10) 67 _ 10
“ 200X 130 200
= 0-67— 0-05=0*62
Example 16. SO c.c. of uaUr contained in the <:<‘^orimclcr a^ifdn i^O
alcohol. Density of watered gm.jc.c.
Volume of the alcohol
Density of alcohol ~
Mass of the alcohol ♦”!=
Volume of water “
Density of water =
Mass of water ’" 2 —
Time taken by alcohol to fall from 40 C
to 1 5V, h=
Time taken by water to fall from 40'’C'
to 15°C, h =
Mass of the calorimeter =
Specific heat of the material of the
calorimeter
= 80 c.c.
=0-8 gm./c.c.
80 X 0*8=64 gms.
80 C.C.
:1 gm./c.c.
sox 1=80 gms.
8 X 60=480 secs.
= 12X60=720 secs
= 150 gms.
= 0*1
water equivalent of the calorimeter u;= 150 X O'l 15 gms.
SPECIFIC HEAT OF SOLIDS AND LIQUIDS
101
Hence
o=
(80+15) 480 _ 15
~ 64 X 720 64 ~ ®
755
Expected Questions
1. (a) Define Calorie, C.H. U. and B. Th. {J. and derive the relation
between them. {A.M.I.K.)
(It) Define specific heat and thermal capacity. How docs specific
heat differ from thermal capecity ?
2. How can the specific heat of a solid be determined ? Describe an experi-
ment to determine the specific heat of a solid. Mention the various sources of
error and how arc they avoided. M I E )
+v.« ® calorimeter, how would vou use it to determine
the specific heat of a metal or liquid available in a small quantity.
a solid'at Pctrtem%ratuTes“"‘° ol
liquid^' ^ simple laboratory method to find the specific heat of a
it is likeIv^*o hnM out the conditions under which
by this meSod ^‘ of determining the specific heat of a liquid
of a liq’uid*'’’ flow molhod of nieasuring tho specie hra!
flow .etflS.
CHAPTER V
SPECIFIC HEAT OF GASES
5.1. Two specific heats. While defining specific heat of a solid or a
liquid it was assumed that whole of the heat acquired by a substance
is used for raising its temperature only.
If the specific heat is defined merely as the quantity of heat
required to raise unit mass through 1°C, then its value will depend
not only on the nature of the substance, but also on the amount of
external work done due to the expansion of the substance caused
by the rise in temperature. In the case of solids and liquids this
change of volume is very small and hence the external work done
during the change in temperature is negligible but in the case of gases
large changes occur both in pressure as well as in volume. For this
reason the external conditions under which the heating of a gas takes
place are important and must be stated when specific heat is to be
determined. Generally all measurements of heat for solids and
liquids are made by heating the substance at constant pressure. So
specific heat of solids and liquids is always found at constant
pressure. But a gas can have any value of specific heat lying
between zero and infinity depending upon the manner in which it is
being heated.
Thus when a gas is heated, ordinarily there is an increase in
volume as well as pressure in addition to the rise of temperature.
For the sake of simplicity either the volume or the pressure may be
kept constant. Therefore, a gas has two specific heats
(f) Specific heat at constant volume (C^) i.e., when the gas is
heated at constant volume.
(ii) The specific heat at constant pressure {Cp) i.e., when the gas
is heated at constant pressure.
These two specific heats are called the principal specific heats of
a gas.
5.2. Specific heat of a gas at constant volume.
heat required to raise the tiwperaivre
when its volume is kept constant. It %8 denoted by c*.
It is the amount of
of a gas through 1^0
102
SPECIFIC HEAT OF GASES
103
Let 1 gm. of a gas be enclosed in an air-tight container of
capacity V c.c. When the enclosed gas is heated, its temperature
goes on increasing resulting in an increase of its pressure at constant
volume. Whole of the heat absorbed by the gas is used in raising
the temperature. Let i gm. of enclosed gas require q units of heat for
raising its temperature through l"C.
then ^=mass x sp. heat at constant volume x rise in temp.
== 1 X Cy X 1 —Cy
Hence for m gm. of gas and for temperature rise of d9 = (02 — 0i)»
the total quantity of heat ^ is
Q^^inCy {^2— 0i)
—mCydd (i)
5.3. Specific heat of a gas at constant pressure. It Is the amount of
heat required to raise the tzmperatun of one gram of a gas through
l^C when its pressure is kept constant. It is denoted hy Cp.
Let 1 gm. of the same gas be enclosed in a cylinder fitted with
frictionless movable piston. On heating, the enclosed gas e.\pands
thus pushing the piston upwards against external or atmospheric
pressure which remains constant. For pushing the piston upwards
some energy is required.
Let 5 ' be the amount of heat required by i gm. of enclosed gas
for heating it through at constant pressure, then
g'=lXCpXl = Cp ...{ii)
This amount of heat is utilised for the following two purposes : —
(i) For raising the temperature of the enclosed gas through rC
This heat q units remains within the body of the gas and is tlie same
as explained in relation (t),
(n) For doing the external work. Some heat energy of o, units is
required for enabling the gas to expand against external pressure.
If P is the external pressure and dV the increase in volume then,
I^ d\^
gi = — r-
or
or
=thermal equivalent of work done.
1 X Cp X I ^ I X Cy X 1 “F
fh. Th^us the specific heat at constant pressure is always neater
than the specifc heat at constant volume by the thermal equivalent
of work done -j- against external pressure.
If m gm. of the gas are heated from e°C to 0 und^r *
pressure then total heat taken is me )b \ ja constant
been explained above that during heating at^conitant
work has to be done. ^ neating at constant pressure, extra
ENGINEERING PHYSICS
In general we define specific heat corresponding to gram molecule
of a gas and not a unit mass of it. Such specific heats are called
gram-molecular or molar specific heats. The gram-molecular specific
heat at a constant pressure is denoted by Cp and gram molecular
specific heat at constant volume by Cy. If M is the molecular wt.
of the gas then
Cf,=Mcp
and Cy=J/Cy
Hence when heat energy is supplied to a gas, in general part of
it is used in doing some external work and the rest remains within
the body of the gas for raising its temperature. The first part for
doing external work represents the External energy of the gas and
the second part which is utilized for raising the temperature is known
as the Internal energy of the gas. The symbol for internal energy
is E.
5.4. Relation between two specific heats- Consider unit mass
(gm or kg.) of a gas enclosed in a cylinder fitted with a frictionless
movable piston, at a pressure P and absolute temperature 2\. Let it
be heated and the temperature of the gas rises from to absolute.
If the volume is kept constant by increasing
the pressure, then whole of heat is utilized
to increase the internal energy of the gas.
The heat required to raise the tempera-
ture by {T2—7'i)=rfT at constant volume is
given by
g=lxCy dr=Cy dT
If instead of the volume, the pressure is
kept constant and the volume is allowed to
increase, then an additional amount of heat
equivalent to the work done by the gas is
also required. This work done is calculated
as follows.
Let A be the area of the piston which moves outwards by distance
dx due to the expansion of the gas. then
The force on the piston=Px-4
Work done by the gas in expanding against the external pressure
=Force X distance moved
=PxAy.dx
=PdV
where dV is equal to the increase in volume. =.<4 xdx
Additional amount ofheat required
where J is the Mechanical equivalent of heat.
T'/^ro/v
vcyunoer
Fig. 5.1
SPECIFIC HEAT Or GASES
105
If Cp is tlic specific heat at constant pressure, then total heat
required to raise the temperature through dT is equal to CpdT
!Z' = ?-fry,
CpdT^C,,dT + ^j'-
or
{Cp-C,,)dr^
PdV
J
(0
But for a perfect gas PV^RT
Differenting we get (pressure P and R are constant)
PdV=RdT
Substituting this value in equation (Z) we have
{Cp~C^)dT=
n the gas constant R is expressed in heat units then the equation
(Cp— Cu)=-— can be written as
or
Cp~C
V
:R
constant important since it proves that the charade, islic
specific hfats.“ '^‘‘^^mnee between its two principal
is cal^ie^" ' is erg. unit of heat
then
R
cal/gm/°iC.
is KTlo!car‘' jo'-le and unit of heat
then
Cp— Cv=-^Kilo-cal./Kgy®A'.
.o .."ufs,r.£
Cp Cl,5=S
R
specififhea^songlTisTve®^ two principal
is denoted by y (gamma/ constant in thermodynamics
y y iaamnia). It is also knoivn as adiabatic index.
106
engineering physics
Cp
Since Cp is always greater than C^ therefore the value of y is
always greater than unity t.c., y >l
Now Cp=:Cy-f-
where i? is gas constant in work units.
y==-^^=
Cy-f-
R
or
y-1
Cy Cy
B
JC^
Hence
y-l =
Cy-
R
JCv
R
AR
1
where A — -j—0'24 cal/Joule
Also Cp
the equations become
Cy-
and
J(y— 1)
(y— 1)
:yCy
yAR
(y-0
molar specific heats
R
' J(y— 1)“
(y-1)
vR
yAR
5.6. Change in Internal Energy of a gas- When m gm. of a gas are
heated 'from /.V to t/C the increase inits internal
energy dE is given by
dE=mCv{T^-'^i)
AR
But
Cy—
•
• •
dE=-
Also
and
#
P^Fz-PiVi^^
or
mR {Tz — 2'2) —
— 1
...(»)
— 1
SrcCIFIC HEAT OF GASES
107
Hence putting this value in equation (i), we have
dh=^- — in heat units
y— 1
^ — in work units
y — 1
If gm.-mole of a gas is considered, then the equation (t) becomes
{T^~T,) ^RA (T^-T,)
y — 1 y — l
If R is in heat units then,
d£=Rl?V:^)
y—l
Example l. Calculate the specific heat at constant volume for air,
given that specific heat at constant pressure^O'ZS. Density of air at 27‘'C'
07^ standard atmospheric pressure=^ 1-18 gmsjlitre and J=4‘2 joulcsjcal.
Given that g~981 cmjsec.^ and density of mercury is 13-6 gmsjc.c. (P. !/.)■
Mass of 1 litre of air = i*i8 gms.
Pressure of air P=76 cms. of Hg.
= 76X 13*6X981 dynes/sq. cm.
= 1*013 X 10® dynes/sq.cm.
Temperature of air 2'c=273 + 27 = 300'’A'.
PV^mRT
or
m
papier =ppr
or
P=
pT
where P— density of air=:mass per unit volume in gms. per c.c„
1*18
substituting the values we have
1*013 X 10®
^ 1*18
or
R_
J ^
But Cp— Ct,=
X300
1000
1013 X 10 6
1*18
X300
X
1
1000
R
4*2 X 10’
.-;:^= 0 * 0682 l
108
ENGINEERING PHYSICS
or Ct’=specific heat per gm. at constant volume—c
R
P J
= 0*23—0.06821
= 0*1618
Example 2. The specific heal of Argon at constant pressure is
and th-' ratio of its specific heats is 1 667. Calculate the mechanical
equivalent of heat. One litre of argon at O'^C and 76 cm, pressure weighs
17S6gm. Given that g—USl cm. .'sec- and density of mercury^lS'd
gms c.c.
According to general gas equation
py = mRT
cury
(A.M.LE.)
m
or P=^BT = pRT
I
i?=
PT
given that at X.T.P.
P = 76 X 13*6X 981 dynes/sq. cm.
= 1013X10® dynes/cm-
T = 2m°K
= 1-786X 10"® gm/cm®
m 1*786
1000
Substituting the values in relation li), we have
i?=
1*013 X 10® =2 08X 10® ergs/gm/®X.
1*786 X 10-®X273
or
Given Cp=0-127 and ££. = 1.667
1*667
R
Now Cp~-Cr;= j-
or
J=
R
2*08 X 10®
Cp — C
“0*127 — 0*077
=4*16X 10^ ergs/cal.
=4 16 joules/cal.
Example 3. If one litre of Hydrogerr ^eighe 3-09 gm.
...nd if its specific heat at constant pressure ^s 3 40.1. Jma
andcvr J=4-18jouleslcal.
Now —mRT
inRT
•a
or
V
pRT
or
R=
pT
SPECIFIC HEAT OF GASES
At N.T.P.
Pressure P=1’013 x 10® dynes/cm-
and temp. T=2ll^K,
009
To^
Now
1013 X 10®
0*09
X 27 3
1000
J
=4-12 joules/gm/®Ar
or
li 4- 12
<'v=c»— -T- ~ 3-409 — 3-409 — 0*9856
^ 4 * 1 8
I0<>
— 2-4234
Cp 3-^09
2’’423'4
Example 4. Calculate the specific heat at constant volume for
hydrogen, given that its specific heat at constant pressure is 6‘S5 cals/gm
imlecule. Density of hydrogen at N.T.P. is 0-0899 gmllitre and
joulesfcal.
The gram molecular specific heat of hydrogen at constant
pressure Cj,= 6'85 cals/grn mol.
Density of hydrogen at N.T.P.=0*0899 gm/litre.
The molecular weight of Hz is 2.
Volume occupied by one grn. molecule of gas at N.T.P.
1000
0-0899
X2==22430 c.c.
Pressure at N.T.P. = 76 x 1 3-9 x 981 dynes/cm- = 1-031 x 10 ®
Temperature at N.T.P. = 0®C=273®A’.
^PPiying the equation
J 03IX10® 1000X2
T 273 ^ 6o899
dynes/cm^
=8-318 joules/gm. mol/®A.
Specific heat at constant volume C'„=Cp—
R
J
= 6-85 —
8-318
4.2
= 6-865 — 1-985
=4*865 cals/gms mo!.
110
ENGINEERING PHYSICS
Example 5. Calculate the difference in two specific heats of one gm.
of Helium. Molecular weight of Helium^d.
Mol. \vt. of Helium=4
one gm. mol, of every gas at N. T. P. occupies 22.4 litres
=22400 c. c.
Hence volume of 4 gnis of Helium at N. T. P . =22400 c. c.
Volume of I gm of Helium at N.T.P.=?^^^=5600 c.c.
4
Pressure Ps=76 cms of Hg.
Temperature T=0°C’=273®A',
Hence i? X 5600=2*078 jules/gm/1® K
Also J=4*2 joules/cal.
^ R 2*078 «
But Cp—c^^-j- — — — =0*3 approx.
5.7. Joly’s dififereDtial Steam Calorimeter for the determination of specP
fic heat at constant volume. This apparatus was devised by Prof. Joly
for measuring the specific heat of a gas at constant volume. This method
depends upon the principle that heat lost by hot substance is equal
to the heat absorbed by cold substance.In this method steam is passed
to raise the temperature of the enclosed gas at constant volume. As
steam loses heat, it condenses. The heat lost by steam while
condensing is equal to the amount of heat gained by the gas,
The apparatus consists of a double walled metal enclosure M
known as steam chamber. It is placed below a sensitive balance B.
Two exactly identical hollow copper spheres h^ and 1u of about 6*7 cm
in dia. and capable of withstanding a pressure of about 35—40 atmos-
pheres are freely suspended from either arm of the balance, by fine
wires of platinum pp. Each sphere is provided with a small tray 1 1
at the bottom known as “catch water” to hold the condensed steam
and is provided with a “shelter ’ S above it so that steam condensed
on the ceiling of the chamber may not fall on the sphere or catch
water. The holes H. H. through which wires pp carr}-ing spheres enter
the steam chamber are made narrow so that any steam escaping
through them may not disturbe the weighing and are lined with
plaster of Paris for preventing the formation of water drops at the
holes so that these do not cause an error in weighing. The formation of
water drops due to the condensation of steam on the suspension wires,
in the holes, is further prevented by placing small heating coils c. c.
around the suspension wires as showm in Fig. 5.2. These coils are
heated by passing an electric current and thus prev’ent the condensa-
ion of steam.
SPECIFIC HEAT OF GASES
111
spheres are thoroughly cleaned from inside and evacua-
ted. They are then counterpoised on the balance. Now one of the
spheres say Aj.is exahused and the sphere is filled with the gas under
test at a high pressure. It is again counterpoised against, the
empty sphere and thus the mass of the gas in is found out. Let it
oe m gms. The'- initial temperature ty of the steam chamber is
i«f ^ ^ thermometer. Now dry steam is admitted
W chamber through the inlet hole 1 at the back and
^ bottom at 0.0. The steam starts condensing on botli the
p eres until they come to the temperature of the steam. More of
sieam condenses on the sphere than on the sphere because in
r<»arV.^!f ^ ^Vhen steady condition is
Jk+ further condensation of steam occurs, additional
added in the pan attached to sphere until the two
spheres are again counterpoised.
thA give the weight w of the extra steam condensed on
of uiicd sphere to heat the contained gas to the temperature
ar.^ ii ^ .^^^Perature of steam is also noted. Both the spheres
actiy similar and so they have same thermal capacities.
specific heat of the gas at constant volume and L
Mie latent heat of steam, then
Heat gained by the gas== 7 ft cals
Heat lost by stean =wL
But Heat gained =Heat lost
or
{h-h)=wL
ivL
.applied.
Mh—h)
For getting accurate result the following corrections have
to be
112
ENGINEERING PHYSICS
1. The first weighing is taken in air at atmospheric pressure at
ti^C whereas the second weighing is made in steam at t^^C. Hence it is
necessary to reduce both the weigliings to vacuum to avoid error due
to buovanev.
2. It is impossible to get two spheres of exactly same size and
thermal capacity. To avoid the error due to this factor, repeat the
experiment by evacuating ho and filling with same gas at same
pressure and temperature. Calculate Cy in this case. The mean
value of botli the specific heats will give correct value.
3. The sphere containing the gas expands, botli due to the rise
in its temperature and also due to increased pressure of the enclosed
gas. Some work is thus done against external pressure so that the
calculated value of Cy is rather too high. Correction has to be applied
for it.
4 . Correction for the change in buoyancy arising from the
increased volume of the sphere must also he applied.
Example 6. I n J oly' s steam calorimeter the dijference in weights of
the steam condensed when the copper sphere is filled with the gas at 16
atmospheres pressure and when exhausted is 0’3S gm. U
the gas contained in the sjyhere is 1000 c.c. and its density is O'OOS
gm.x.c., calculate the specific heat at constant volume. 2 he initial
temperature of the gas is 16^C and the temp, of the steam equal to 100 C.
Latent of steam is 620 cals.jgm.
• «
\'olume of the gas
Density of the gas
mass of the gas=volume x density M
= 1000 c.c.
=0'008 gm./c.c
= 1000X0-008
= 8 gm.
Initial temperature
Temperature of the steam
Latent heat of steam
mass of steam condensed
<1=1 5°C
(2=100‘’C
L = 540 cals./gm.
7rt=0-38 gm.
Let Cy be the specific heat of the gas at constant volume.
Hence applying the formula we ha\e
mL 0-38 X 540
Cv^
M {.h-h) ’ 8000 - 15 )
= 0 3
5.8. Determination of Specific heat a^^onstant pr^ure Renault’s
exoeriment An accurate method of measuring specific heat at 9 °"stan
pr^ssTe wafSivised by Regnault. It depends upon the principle that
SPECIFIC HfAT OF GASES
113
heat lost by hot substance is equal to heat gained bv cold snhst'inr^
gas
AltTlr^ surrounding it by a large water tank as shown n Fig 5 ^
A thermometer T, measures the temperature of water in the tan^
mea‘sur:d%W?,fa
opena%h®:^ta^nd“la\l"g thr-^^
— -o*
long so that the gas while nassi^ * h* u ^ u® ‘® sufficiently
temperature of thi oil Ch After® attains the
?as is made to pass throughlhe tXe the spirals! the
‘0 a calorimeter D containing- wtrl ■^PP"'' ^P'^^l placed
114
ENGINEERING PHYSICS
In order to avoid the loss of heat by the gas in passing
from the bath to the calorimeter, the calorimeter is
placed inside a double walled vessel filled with water to minimise
?he loss of heat due to radiation etc. In order that the ?as “ay
enter the calorimeter at the temperature of the oil toth ‘t m
as near the bath as possible. A wooden screen -IF
between oil bath and calorimeter so as to minimise error due to
gain oi heat by radiation from oil bath.
Let (i=the temperature of the hot bath
U and /3=the initial and final temperatures of the water in the
calorimeter
w,=water equivalent of the calorimeter and its contents
j^_n^ass of the water in the calorimeter
Heat gained by the calorimeter and its contents
= («.'+ 3 /) (^3-^2) cals.
The temperature of the hot gas falls through {ti fg) ,
during the experiment
7n =mass of the gas escaped
^^^specific heat of gas at constant pressure,
If
and
then
Heat lost by the escaped gas=mCp («i-
Heat gained=Heat lost
^a + ^3
)
or
/ u~\-h
mf<i
™ .1 .h. s.
initial pressure of the gas in the reservoir=P,
Final pressure of the gas in the reservoir 2
SPECIFIC HEAT OF GASES
= Po
=F
ns
Temperature of the gas
Density of the gas at N.T.P.
Volume of the reservoir
^o\v the density of a gas is directly proportional to the pressure
and inversely proportional to the absolute temperature.
Density of the gas at pressure and temperature T
P -P
P.-Po X-p
where and Tq are normal pressure and temperature on the absolute
scale of temperature.
Pi=Po
s,273
X
I 76 ' y.
. . (D
Similarly the density pg fhe pressure is given by
Pi ..273
P. = P0 7|X
T
••.{it)
From (t) and [U), we have
_ Po {Pl-P^) 273
r 1 — r 2 L : 7 :
76X'I'
the mass of the gas
«*={Pi^p2) V
_ Po (Pt-Pj) 273
76 xr ^ ^
Knowing every quantity on the right hand side, m can be calculated
Example 7. In a7i experiment for the determviaiion of the
heat at constant i^ressure hy Regnault's method.
Temperature of the oil hath ty^^250°C
Initial temperature of water m the calorimeter i2^27'^C
Final temperature of water after passing hot air in
the calorimeter ^
Water equivalent of calorimeter and Us conte 7 tts w=10'52 gms
Mass of water ta1ce7i in the calorimeter M=50'43
Constant temperature of the bath containing
the gas holder i^27^C
Density at N.T.P. P^O-001293 g,njc c
Capacity of the gas holder '
Decrease of pressure of air after the experiment P^-P^
^15‘2 cm. of By
116
ENGINEERING PHYSICS
in
Calndate the mass of air passing through the calorimeter and also its
specific heat at constant volume.
The mass of air passing through the calorimeter
_ Po(-^ 1 P 2 ) 27 3 TT
“ 76xr
_ 0 001293 X X
“ 76 X(273-f27)
Specific heat of air at co stant pressure.
^ (»‘ + -l^) (^3 — M
^ f , ^2 + ^3 ^
V‘“' ~^)
(10-52 -1-50-43) (37—27) 60- 95X 10
= r’" 37+271 —11-77X218
n-77 250 J
= 0-237
5 9 Scheel and Huese continuous flow method for the determination of
C;,. Regnault was heated to high
by allowing flow mto i„to another
S insid:' a calorimeter ani
much more accurate i"®*hod o e er determining
in using the same technique g Barne’s continuous flow
(erecTrfc) "method Vht' me^od has been designed by Scheel and
Huese for finding the Cp of a gas.
The apparatus consists of a glass tube of special shape as
shown in the Fig. 5.4. -i T^H+innm
In the innermost part of the tube there is a coil of platinum
WATER
GLAS5TUBE
/
jacket
TO PLATINUM
RESISTANCE
thermometer
Fig. 5.4
wire
H which is sealed into the glass and is heated by connecting the
SPECIFIC HEAT OF GASES
117
leads Z-, and Z -2 to a battery B through a rheostat a key and an
anmeter The potential difference is measured by connecting the
voltmeter V across the leads and
The glass tube is sealed into an outer vacuum jacket to
reduce the loss of heat due to radiation. The gas enters the glass
tube through a long metal tube M in which is scaled a platinum
resistance thermometer Pj which measures its initial temperature. A
similar platinum resistance thermometer is fitted on the other side
to measure the temperature of the out going gas.
To begin the experiment, the apparatus is placed in a constant
temperature bath. A steady flow of gas at constant pressure is
introduced into the metal tube at A and its temperature is measured
by thermometer P^. The gas then flows through spiral tube and
then travels in the direction of the arrows through the zig-zag path
and finally comes in contact with the heating element. The purpose
of providing the zig-zag path is that any heat lost by the heated gas
IS brought back into it by the incoming stream of the gas and there is
Tio wastage.
The gas then passes through copper gauze Q. the function of
which IS to mix the gas thoroughly so that it attains a uniform temp-
•erature. The gas finally flows over a second thermometer and it
issues out at B.
constant rise in tempera-
TSstan^e^+brmn T readings of the two platinum
resistance thermometers at the entrance and exit respectively. Let
these thermometers be (, and t °0
pressures measured by noting the initial and the final
-Lu * reservoir. If V be the reading of the
Itmeter. I that of the ammeter,then heat produced electricafly by the
wire calories
4'2
When the current flows for t seconds
Heat taken up by the gas=m(7y {k~h)
Neglecting losses we have
mCp
or =
Vlt
4 •2m
and hen“‘clrbe iegle^rted f or" aU pracUc^^pur^os*^*'”"
as it U^produetd e^cVlL^fy* ®®''®‘'ated can be measured accurate!
118
ENGINEERING PHYSICS
(m) Since the temperatures are constant the thermal capacity
of the apparatus can be neglected. Moreover, steady temperatures
can be read \vith greater accuracy' than the varying temperatures.
(r*) The method can be used to study the variations of specific
heat with temperature. To find the specific heat at very low tempe-
ratures, the gas is initially passed through a low temperature bath in
which the whole apparatus is also immersed. In case of helium and
other rare gases, Scheel and Huese further modified the apparatus so
as to have a closed circuit.
Expected Questions-
1. (a) How would you account for the two specific heats of a gas ? (A.M.I.E.)
(6) Explain whv the specific heat of a gas at constant pressure is ^^ter
than at constant volume, obtain an expression for the difference between the vajues
for a perfect gas.
2. Describe an accurate method of measuring the specific heat of a gas at
constant pressure. j c
3. Describe Joly s steam calorimeter and explain how it may be used or
determining the specific heat of a gas at constant volume.
4. Give an account of the continuous fi(>w method of measuring the specific
heat of a gas at constant pressure and point out its advantages.
5. What is meant by specific heat ? How does the change of the temperature
scale effect the numarical values.
CHAPTER VI
THE MECHANICAL EQUIVALENT OF HEAT
6.1. Thermodynamics. It is the scicywc that disaisscs the relation of
heat to mechanical work. It establishes the equivalence between work
done and the heat produced.
6.2. First law of Thermodynamics. According to this law, when a
given amount of woik is done, an equivalent amount of heat is
produced. When a certain quantity of heat disappears, an equivalent
amount of work is always obtained. Thus we are perfectly justified
in saying that besides kinetic and potential energies of matter there
is another form of energy called heat energy. The relation between
work done and heat produced is expressed in the First law of
thermodynamics which is stated after Maxwell as follows.
* 'When, mechanical work is completely cojiverted into ^ heat or heat
is completely converted into mechanical work, then for each unit of work
which is converted into heat, a definite quantity of heat is produced and for
each unit of heal converted into work a definite quantify of work is
furnished.**
Or
"Whenever work is transformed into heat or heat into work, the
quantity of work is mechanically equivalent to the quantity of heat.*’
If W is the amount of mechanical w'ork which when completely
converted into heat produces H units of heat, then it is found that
TFocH
TF
or —ssconstant
u
The constant is known as Joule's mechanical equivalent of heat
and is denoted by
If H=unit quantity of heat then J^W. Thus Joule's mechanical
equivalent of heat may be defined as the amount of work which must
be done to produce a unit quantity of heat.
119
120
ENGINEERING PHYSICS
Units of J.
On C,G.S. system of units the work W is measured in ergs and heat
II in calories. The value of J on this system is 4*2 x 10’ ergs per calorie.
or J =4*2 X 1 0’ ergs per cal.
On M.K.S. sj^stem of units the work W is measured in Joule’s and
heat H in calories or kilo-calories.
The value of J=4‘2 Joules/calorie
=4200 Joules/kilo-calorie
Also if ir is measured in Kg. -metre and H is in Kilo. cal. then the
4200
value of J nietre/kilo-cal. or 426-8 kg. metre/kilo-cal.
Note. 1 Kilo cal. =426*7 kg.-metre and l kg. meter= 9.81 joules.
Example 1. A water fall is 420 m. high. Assuming that no energy
is lost by water, find how much hotter the water at the battom is than that
at the top J=4'2 Joulesjcal. and g=9'81 mjsec^.
Tet m kg, of water fall through a hight h and the temperature
rises through
Work done 'vhen m kg. fall through a height h
W =mgh Joules.
=mx9*81 X420 Jdules.
Specific heat of water S= 1
If H is the heat generated, then
H=^mSt—mt Kilo-cal.
But \V=JH.
where J=4*2 Joules/cal.
=4200 Joules/Kilo-cal.
.*. mX9*8l X420 = 4200XmXt.
or
S- 9*81 X420
4200
Example 2- A lead hulltt at a temperature of 4TC strikes against
an unyielding target. If the heat produced by sudden ^loppage %s gu^t
sufficient to melt the bullet, with what velocity does the bullet strike the
target. It is assumed that all the heat is produced withm the bullet.
Melting point of lead=327°C, specific heat of lead^O OS caljgmrC
Latent heat of fusion of lead— 6 caljgm. J=4'2 Joulesjcal.
THE MECHANICAL EQUIVALENT OF HEAT
121
Let m kg. be the mass of the bullet.
Heat required to raise the temperature of the bullet from 74"C
to 327°C=mxO’03X(327--47) = 8'4 m kilo-cals. —8400m cals.
Heat required for melting^m x 6 = 6m kilo-cal. =6000 m cals.
Total heat required for melting //=(8'4m+6j«)
= 14‘4m kilo-cals. =14400 m cals.
Kinetic energy of the bullet ir = imy- joules,
where v is the velocity in metres/sec.
But W=JH.
^»H-^ = 4*2X 14400m
or r2^4*2X2X 14400
r = V2^800 X4-2 = 348 m/sec-
Example 3. From what height must a block of ice drop in order that
it may be melted completely. It is assumed that o0% of the energy of fall
is retained by the ice.
Let m be the mass of ice and h the height from which it is
dropped, then
Amount of work done W =»n^^=mx 980 X h ergs.
=mx 980x^X 10~’ joules.
Energy utilized to produce heat=-Y^ xmx/tx980 x 10“’ joules
Heat required to melt m gms of ice H=mx 80 cals.
mX80= ix «». x7.x 980X 10-^
4-2
or
h
80X4-2X 10X2
980
= 6860000 cms.
=68*6 K. metres
Example 4. A closed caid hoard tube one metre long contains some
lead $^8 at one end. The tube is quickly inverted so that shots fall
mroyhthe whole length of the tube. The operation is repeated 100 times.
7h temperatnre if the specific heat of lead is 0 03 •
ta/ce g^9'81 m.fsec^. (-P )
Let the mass of lead shots=m kgm.
122
ENGINEERING PHYSICS
Work done when the shots fall lOO times in the card board tube
tlirough a height of i metre
If z=n mgh joules
= 100xmx9*8] XI joules.
=981 m joules
J temperature of lead shots of specific heat
0 03, then ^
Heat produced =m5^
=m X 0*03 X t kilo-cals.
TIic card-board is a bad conductor and does not take up any
heat.
Now W=JH
where t/=4200 Joules/kil-cal.
981 ?/i=4200XmX0*03 X/
or
981
4200X0*03
Example 5. A feed pump delivers 1200 litres of water per hour
against a constant head of 100 metres. What is the work done per hour in
heat %inits.
Quantity of water raised per hour=i200 litres = l200 kg.
Work done by pump per hour s=Mgh
= 1200X100 kg. metre
Now 1 Kilo-cal. =426*7 kg. metre
Work done in kg. metre
.*. Work done per hour m heat units= 426^7
Kilo-cals.
I200XJ00
426*7
= 281*3 kilo-cals.
6.3. Determination of Mechanical equivalent of Heat. The following
methods are commonly used for determining the value of J.
1. Joule's E.xperiment.
2. Rowland’s Experiment.
3. Calender and Barne's continuous flow method.
4. Callender's Brake Band Method.
5. Electrical method and
6. Searle's Friction Cone method.
123
THE MACHINICAL EQUIVALENT OF HEAT
(1) Joule’s experiment for finding J. The apparatus used by
Joule is shown in Fi^ 6.1. It consists of a large calonmcter C to
the sides of which are soldered vanes F I . The calorimeter is
provided with a water-tight lid having a hole m its ccntie through
Fig. 6 1.
which passes the spindle S. The spindle carries paddles PP which can
rotate in between the corresponding openings in the fixed vanes.
The vanes prevent the rotation of water along witli the revolving
paddles. The spindle S is connected with a removable pin E to a
drum D and can be connected or disconnected from the
drum by fixing or removing the pm when desired. Two coerds are
wound round the drum in such a way that their free ends leave it in
the opposite directions but rotate the drum in the same direction. These
cords pass over friction less pulleys P^ and P^ and to their free' ends
are attached two equal weights W.W. A handle is connected at th top
of the drum. The vertical distance through which the weights fall is
measured with the help of scales placed on both sides of the apparatus.
A known mass of water is put in the calorimeter and its temperature
is recorded udth a sensitive thermometer T placed in it.
When the weights are allowed to -fall the spindle turns and the
paddles produce a churning motion of the water inside the calori-
meter. The potential energy of the weights is converted into the
kinetic energy of water. The motion of water is prevented by
vanes and thus the kinetic energy of water is converted into heat
^ I- pi" ^ is removed,
the spindle is disconnected and the weights are brought back to their
mitial position by rotating the drum with the help of the handle.
The pm £ IS once again fixed and the weights are allow'ed to fall
as before. This process is repeated a number of times and the tempera-
J24
ENGINEERING PHYSICS
caloriometer is once again recorded. The difference of the
t«o readings give the rise in temperature of calorimeter and its
Contents.
Let m be the mass of water in gms,, ic the water equivalent of
calorimeter, vanes, spindle and paddles and t be the rise in
temperature.
Heat produced H ={m-\-w) t calories
. gms. be the mass of each weight falling n times through a
iieight h cms. during the experiment,
Work done \Y=2Mgh x n ergs
Assuming that no heat is lost by radiation and other causes, the
mechanical equivalent of heat,
^ _ 2 Mgh X n
H (mH-te) t
As the work is measured in ergs and heat in calories, the value
of J on the C.G.S. system expressed in ergs per calorie. If w and w
are measured in kilo-gms, H will be in kilo-cals. If M is in kilo-gms, h
in metres and g in metres per sec^ then W is in joules. Hence the value
of J in M.K.S. system is measured in joules per kilo-cal.
Sources of error. This method does not give accurate results due
to following errors : —
(i) Loss of heat by radiation.
{ii) Loss of energy due to friction of pulleys.
{Hi) Kinetic energy of the falling weight. The whole of the
potential energy of the falling weights is not converted into work but
a part of it is utilised in giving a kinetic energv to these weights
themselves. If V be the velocity of each weight as it reaches the
ground, the kinetic energy of each weight in one fall=| il/F*, the
kinetic energy of both weights in one fall,
= 1 MV^x2=3fV’^
The kinetic energy of both weights in n falls
— nJilV^
.*. Net work done \V^ 2 Mghn — n3IV^
Hence
{ 2 Mgh—MV^)n
O — —
{m'^w)i
Hxample 6. In Joule s experiment a mass of 1.45 k.gms. was^ allowed
to fall through a height of 30 cms. The process was repeated 200 ^
the rise of temerature observed was 2.0°C. If the thermal capcity of
-calorimeter and its contents is 100, find the value of J.
THE MECHANICAL EQUIVALENT OF HEAT
125
Let M be the mass in gms. falling ii times through a height h
during the experiment.
d/=l'42 kgms. = 1420 gms.
Height /t=30 cms.
No. of falls 71=200
Work done by the falling mass,
W —Mgh X n ergs
= 1420X980 X 30X200 ergs
= 1420X980 X 30 X 200 X 10"' joulcs
Thermal capacity of the calorimeter and its contents
= 100
Rise in temperature t=2'^C
Heat produced // = 100 x 2=200 cals.
Hence
W _1420 X 980 X 30 X 200 X 10~’
'H ~ 200
= 418 joules/calorie.
(2J Rowland's Experiment. It is an improvement of Joule's method.
Rowland tried to remove the following weak points in Joule’s
determinati6n of mechanical equivalent ot heat.
(t) The rise of temperature was small.
(n) The mercury thermometer was not compared with a standard
thermometer.
(m) It was assumed that the specific heat of water between
0 — ioo®Cwas constant.
Description. The apparatus consists of a calorimeter C fitted
with a set of vanes VV. A vertical shaft AB is suspended by a torsion
wire T from a torsion head H. A set of paddles P,P attached to the
shaft is made to rotate inside the calorimeter by means of an axle
passing from below and connected to a steam engine. The number
of rotations made by the axle is recorded by a revolution counter.
The friction of water tends to make the calorimeter rotate along
with the paddles and thus exerts a couple on the shaft AB. The
paddles and the vanes have a very large number of holes drilled in
them in order to make their churning action more thorough. The
calorimeter contains a known mass of water and its temperature is
measured by a sensitive thermometer T placed inside the perforated
carring the paddles. The calorimeter C is surrounded by a water
Jacket to measure the loss of heat due to radiation.
The shaft AB passes through the centre of a circular disc -D to
which it is firmly attached. Two silk cords are wound round the dis«
126
ENGINEERING PHYSICS
in such a manner that tliey leave it tangentially at the opposite ends
Fig 6.2.
•
of a diameter. The cords then pass over pulleys and P^ and are
attached to two equal weights W.W so that the arrangement exerts a
coupletending to rotate the disc and balance the couple acting on
the shaft AB due to the rotation of the paddles. A horizontal arm
r is attached to the vertical shaft AB and carries sliding weights W^.
and FT which can be moved in or out so as to vary the moment of
inertia of the suspended part.
The initial temperature of water in the calorimeter is noted
and the paddles are set rotating. The couple applied on the calorie-
meter by the engine is equal and opposite totlie couple applied by
the weildits WW. The weights TFl^ are adjusted so as to keep the
calorimeter in its equilibrium position. The rotation is continued
dU a “nient rise of temperature has been obtained. The number
of revolutions are recorded and temperature read at frequent
fnten'Ils In this way a number of observations are taken for various
rise of temperatures.^ The value of J can be calculated from any pair
of observations.
If d is the diameter of the torsion wheel D, then
Momment of the couple=T/sfX(f.
If n is the total number of rotations made during the experiment
by the splinde carrying the paddles, then work done
THE MECHANICAL EQUIVALENT OF HEAT
127
Tr=[Moment of the couple] x [angle turned in radians/sec.]
= Mgd X 27rn
If m is the mass of the water and w the water equivalent ol the
calorimeter, paddles and vanes etc. and (< 2 — fi) temperature
then the heat absorbed.
T7 X 27rn
If is in gms, d in cms. and g in cms. per sec^, then W is in ergs.
If m and w are in gms, then H is in calories. Therefore on the C.6-’..S'.
system J is measured in ergs per calorie.
If M is in K.gms, d in metres and g in metres per sec-, then W is
in Joules. If m and w are in K-gms, then H is in kilo-cals. Therefore
on the JI.K.S. system J is measured in Joules per kilo*cal.
(3) Callender and Barne's Continuous flow Method. The method
has already been fully explained in the Chapter on Specific Heat. The
onlj’ differerence lies in the fact that instead of finding the specific
heat,^^ the value of J is found and from the two equations
given in Art. 4.9, we have
mS{t2 — — J — (*)
and TTi'SiJz —
VTt
(ii)
Subtracting (i) from (u), we have.
or
{vi-v'iy
. If F and I are measured in volts and amperes respectively then J
is in Joules/cal. r ^ j
Examle 7. In an experimcyit by contuiitous flotv mcthed.when the rate of
Jtow of water was 11 gm. per minute the heaiiwj current and potential differ^
® ^ amperes and 1 volt and the rise in
« mperafure of the water was 2.5®C.On increasing the rate of flow to 25 4
gm/mmute, the heating current to 3 amperes and P.D to 1.51 volts the rise to
temperature was still the same. Determine the value of J.
F=i volt and F'=l-51 volts
/=2 Amp. and r=:3Amp.
m=il gmsfminute and m'=25*4 gmjminute
<8— <i=:2-5V.
128
ENGINEERING PHYSICS
Now
T7-r7'
A
_ (1*51 X 3 — 1 X2)
‘ (25-4'-^U)(2^ =4*2 joules/cal
(4) Callender's Brake Band Method,
measuring J in the laboratory by a
meclianical method was devised by
Callendar and is shown in the Fg. 6.3.
The copper calorimeter which is
in tlie form of a brass drum D contains
a small quantity of water whose
temperature is measured by thermo-
meter A passing through a hole in the
middle of D. A silk belt T is wound
tightly round the drum, and a weight
ir^ {M(j) is attached to one end,
wliile the other end of the silk belt is
connected to the hook of a spring
balance S. The drum is rotated at a
steady speed by an electric motor
simple method of
T
rO/^fV/NG
1 _ Moro/i
or
Fig. 6.3.
by other mechanical means
against the frictional force between silk belt T and the drum D. The
number of revolutions described in a given time is determined from
the reading on the revolution counter.
During the experiment the drum D is rotated at a steady speed,
in this case the reading on S is fairly steady and \Vt remains
constantly in position. The silk belt also remains fixed in position,
and work is therefore done against the frictional force between the
belt and the drum by the motor. When a suitable rise in temperature is
obtained the rotation of the drum is stopped and the true temperature
rise is obtained.
Load on the silk belt
Tension in spring balance
Water equivalent of drum
Diameter of drum
Rise in temperature
No. of revolutions
mass of water in the drum
The frictional force
{Mg)
S=f=mig
~w
d=2r
=rc
=n.
F^Mg—f=Mg—mig
^{M—mi)g
Note. 1. [The weight Mg acts downwards on the silk belt
the tension acts upward on the belt.]
Work done against friction=Force x distance
* Work doneir =F X l-rzrn
and
the MECHANICAL EQUIVALENT OF HEAT
129
Note 2 . [In one revolution a point on the edge of the drum turns
a distance 2nr.]
Work done
Heat produced
W ~ 2 nrn {3Ig—f)
= 27rrw {Mg — mig)
= 27zm {M —ini)g
_ 2 nm(M ~mi )g
~m,) , , .
= — — 77 —^ ergs'calone
If W is measured in Joules and H^in Kilo-cals, then J in measured
in Joules/ Kilo-cal.
The apparatus is used extensively in laboratories for demonstration
and can give a reasonably accurate value of J provided the following
precautions are observed.
(t) Correction should be applied for loss of heat due to radiation
by taking the temperature of the water and cylinder in the beginning
of the experiment as much below the room temperature as the final
temperature will be above it.
(ii) The surface of the cylinder and the silk should be clean so
that the friction is reasonably constant throughout the experiment.
(tu) Readings of the balance should be taken at intervals during
the experiment, and if the drum is turned by hand, care should be
exercised that the speed of rotation is constant.
Examples. In a Callendar ' 3 experiment, the cylinder ical coover
calorimeter had a radius of 10 cma. a mass of 300 gms, and coMained 150
grw of water A silk hand loaded with 3 kg, weight passed over its
surface and when the drum was driven at a steady rate of 180 revimt
the average reading on the balance which was attached to the other end of
was 560 gm. If the temperature of the water initially wd
L,- 7 if? was calculate the mechanical
equivalent of heat. Specific heat of copper —0.1.
Force of friction between the band and the drum
= (3/— m) = (3000— 560) 981 dynes
Distance covered in one revolution =: 2 rr
= 2 :t X 10 cms.
Distance covered per mmute=i 80 X 27 rx 10 cms.
WorkdoneinlO niinutes = f3000— 560)981 x 180 X Ztt X 10 ergs
Heat produced = (m+w)f=(i50 + 300X0-l) 36 cals.
130
engineering PHYSICS
Now W^JH
J^WiH
__ (30 0 0 — 5 60)X 981 X 10 X 180 X 277X 10
180X 36
=4-12 X 10' ergs/cal.
=^412 Joules/cal.
(S) Electrical method. Tliis mctliod has already been discussed
in the chapter on specific heat of Iiquid.s.
we know that
_ [vi-vry
— 1 \)
Tf .S’ the specific Iieat of liquid is : known the value of J can be
calculated.
f6) Searle's Friction Cone Method. A simple laboratory method in
which heat is produced by friction between two metal cones is due
to Searle. The apparatus is sh>wn in t.ae Fig. 6.4.
TO MOTOR
Fig. 6.4.
tt consists of two truncated brass cones D and E fitting smoothly
.ne the other. The inner cone D serves as a calor.meter and
contains a known mass of water ^ is a sturdy
rests upon the inner cone D, to which it i V .. ^
pins. A leadweight B is placed on the top of disc ^ to noia
;position. A string attached to the circumferance of the disc passe
THE MECHANICAL EQUIVALENT OF HEAT I3I
a pulley P and is k?pt stretched by a known weight [7 fastened to
Its otlier end. The outer cone E is held rigidly in position by m<-ans
ot non-conducting ebonite pieces attached to the copper or brass
cylinder C which is mounted on a spindle S and can be revolved
by means of an electric motor or engine practically at a uniform
speed. The counter 6' records the number of revolutions given t..
e cylinder C and hence to the outer cone. A sensitive thermo-
meter ? and a stirrer are jilaced inside the inner cone as shown.
When the outer cone rotates rapidly the inner cone tends to m we
in the same direction on account of friction between tlie two I'li^
conti-mnu, rotation of inner cone is prevented by applying t e
opposing couple provided by the weigiit IF attached toUie^tring
passing r.nmd the disc D and over the imil .-y as sho.^r v ^
bec.uisc of tiic friction which takes place between the surfac-s in cont ict
eat .s produced wi,ich in turn Juunts the cones Imll sv.rP
^ynge the circumference of
surf£hfcttacV:ftrcL‘«^/^^^^^^ of the
frictional force between the cones. represent the mean value of
Wi ir- ^Xr^TE^MgR
<iueto;rvrryrdV\1ir^^^ the acceleration
on.tl^TnVe^coi^ °“‘or cone
weight on cone Z) the work d^nr^ one reV?T ®
cone assuming the inner cone to'’:emarat rest = 7 x 21 ^^
siven by" " revolutions, the work done. W is
I7=P X 27zm
accuratdy" the‘ valu7f pJoduct'^f r"T7ve? ^"7 p*’® <letermined
this value, we find the mechanical work done
= 27C71
b, <«.,» tt- h... p,oa.„d
132
ENGINEERING PHYSI (
then heat produced H due to friction is given by
j W _Mg'KR'X2Tzn
H
Example 9. Two brass cones jit one inside the other. The outer cone is
rotated about a vertical axis at 600 r.p.m and the inner cone is heldjixed by
an arm whose length, measured from the axis of rotation, is 15 cm. and to
which horizontal pull of 150 gm. wt. is applied at right angles to the arm.
The total mass of the two cones together is 180 gms and the inner cone
contains 25 gm. of water. Assuming that the initial temperature ie 422^0
and that there are no heat losses, fiiidtthe temperature at the end of 5 minutes.
The value of J =4.2 joulesical. and specific heat of brass=0.1.
No. of revolutions made in 5 minutes
= 600 X 5=3000
Force exerted=I50 gm. wt.=0'l5 kg. wt.
= 0*1.5 X9’81 Newtons
The length of the arm from the axis of rotation = the radius
R of the disc = I5 cms=015 metre
Work done against friction in 5 minutes
W=MgXlnRxn
=0*15 x9'81 X2-X0*I5X3000 Joules
=4150 joules
Hence heat produced — calories
Water equivalent of the two cones
= 180 X 0*1 = 18 gm.
Effective wt. of water m=lS-* 18=43 gms.
If the rise of temperature is eC then the heat produced is mSt
where S is the specific heat of water taken to be unity.
4150
43 X 1 X< =
4*2
4150
4*2X43
.'. Final temperature is=22 4- 23
=45X.
23‘^C
1 .
energy
Expected Questions
(а) Discuss the significance of the expression that heat is a form of
(б) What is meant by the term Mechanical Equivalent of Heat.
2. Explain the superiority of Rowland's experiment over that of Joule s.
3. Describe the Callendar and Barne's continuous of
measurement of the mechanical equivalent of heat and pomt out the ad g
this Friction Cone method of determining the value of J and
indicate the principles underlying this method. _ ♦.v.a value of J.
5 (a) Givein brief the brake band method for determining the val i„g.
(6) Describe a simple electrical method for the determination of the valuta
of J in the laboratory.
CHAPTER Vir
EXPANSION AND COMPRESSION OF GASES
7.1. General gas energy temperature equation.
If heat is supplied to gases, in general there is a change in volume
as well as temperature. Due to the increase of volume, work is done
by the gases and a part of the heat energy supplied is converted into
work which may be called as external energy. The remaining energy,
is utilised to increase the temperature and is known as internal energy'.
Hence according to the law of conversation of energy.
Heat supplied=Gain of internal energy -f External energy
(external work done)
dH==dE+-^ (,)
where J is the mechanical equivalent of heat.
The total heat of a substance is known as enthalpy which is
measur^ by the equation (t) from a known condition. If the gas
is cooled t.e.,if the heat is rejected by the gas then H will be negative.
Work done T7 will be negative if the gas is compressed.
7.2. Isothermal expansion of a gas. In a steam engine, the vapour
expands at constant temperature in a cylinder during one stage of
the action. The same type of gas expansion occurs at stages in
gas engines. When a change in pressure aiul volume of a gas takes place at
constant temperature it is known as isothermal change. (Iso-same, thermal-
temperature). Mechanical engineer requires to study the laws governing
the isothermal expansion and compression of a gas. ®
« isothermal change takes place consider
enclosed m a metal cylinder compressed by a piston
an example of an isothermal change. nsrant. inis is
133
134
ENGINEERING PHYSICS
vervSv’^s’oTlfrt”^'., Pjston in the cylinder is released
\er> sio\M\, SO that the gas expands. Work is done bv the eas
and the energy is taken from the gas itself, which therefore decreases
outddo ^ th l^eat is continously supplied from
con>tant, the process is said to be an isothermal expansion.
Tims in an isothermal change the temperature is kept constant by
adding heat or taking It away from the gas. For a perfect gas an
isothermal change is represented by Boyle's law given by the equation,
PF=consfant
+1 ^ ^ rectangular hj-perbola as shown in the Fig. 7.1. As
le F\ diagram for an isotiiermal change, is a rectanc;ular hyperbola,
the process is also known as hyperbolic process-
Work done by a gas in isothermal expansion. Consider a unit mass
of a perfect gas contained in an insulated cylinder and a perfectly
conducting bottom fitted with a piston. If the piston is allowed to
move slowly outwards the gas expands isothermally from an initial
Fig. 7.1. Fig. 7.2.
volume K.and pressure P, (point a) to final volume V, and pressure P.
(point 6) as shown in the Fig. 7.2. The work done by the gas is given
bvlhe area abV,V,. Consider seme pomt ^ where ti e gas has a
pressure P and a volume of F. Let the gas expand ^ th s stag
by a small amount dF=M xrfa-)- T1 is increase m volume rfT- X^)
is so small that the pressure P is supposed to remain censi ant during
the small expansion. The done by the gas dnr ng the smah
expansion is given by the area of the shaded vertical strip JPCD and
is PdV. . , ,,
. dlV=PxAxdx where u4=Area of cross-section ot tre
piston and £7a:= small distance through which piston moves,
or dW=PdV
EXPANSION AND COMPRESSION OF GASES
135
But
PV=JRT
RT
P=
Hence work done dV .
Hence tlie total work done when the gas expands isothermally
from volume to volume Fj is given by
\dW=nT\^^
0 V,
Further since PiVi^P^V 2
A
Vr
W=RT loge-;-
• 1
r ^
=i27’ logcr where r=ratio of expansion
T'l
• ■
or
r= ST log Iog,p
' 2 -T
2
=P.Filog, loge i
' 1 Pq
R represents gas constant for unit mass of the gas.
For one gm-mol of gas R should be substiuted for R.
This work IF is done due to the heat absorbed from the source
Hence heat absorbed from the source source.
cals.
where J is the mechanical equivalent of heat.
operation. The following are the examples
(i) Fusion of solids at their melting point.
(u) Vaporisation of liquids at their boiling point.
Example 1 . Wkat is the pressure inside a vwtor tyre when 275 litres
air w pumped xrUo it at constunt temperature and nJmalJessure Tht
internal volume of the tyre is J50 litres, pressure. The
Initial volume of air F, =275 litres
Initial normal pressure P^^i atmosphere
136
ENGINEERING PHYSICS
Final volume of air Tg* 150 litres
Final pressure of gas P 2 = •
Now, for an isothesmal process PF=!Constant
or
Pi X 275
P2 =
150
11
6
= I’83 atmospheres.
Examle 2. A gram molecule of a gas at 127°C expands isothermally
until its volume is doubled. Find the amount of work done and heat
absorbed. Given that mechanical equivalent of heat 4.2 joules j calorie and
P=:8 3 douUsjgm. mole.j^^K.
The work done when a gas expands isothermally from
the volume I'l to Vo at constant temperature T is given by the
relation,
and
W^BT loge
for one gram-molecule of the gas.
W=rT
for one gram of the gas.
where the sj-mbols have their usual meanings.
Temperature P=273 + 127=400°^
P=8*3 joules/gm. molecule/® A”
’-1 = 2
F V
loge -j|-= 8-3 rx 2-303 log,„y
* 1 * ^
= 8-3 X 400X2-303 Iog,o2
= 8-3 X 400 X 2-303 X O' 30 10
= 2*3 X 10® joules
* . _2-3Xl0®
jibsorDcd*^ — j“
J 4*z
—548 calories.
Example 3. Calculate the amount of work done when 10
of air at a pressure of 2 kgm. /sq.cm, and at a
compressed isothermally to a pressure of 10 kg./sg. cm- Also ca, cu
heat rejected during the proce.'iss. and the change in interna energy-
Work done during the isothermal operation on the gas.
V P
IF=P,F, loge fT- =PlV, loge ^
Pi=2 X 10^ kg./sq. m.
El =*10 cu. metre
Pj=I0Xl04 kg./sq. m.
expansion and compression of gases
137
Work done =2 x lO^X 10 loge
lOX 10^
2X10*
=321.880 kg. metre
As the temperature of the gas remains constant, so docs the
internal energy. Hence the change in internal energy is zero during
the process.
Heat rejected=Work done
_32I880 _321880
J 426*7
= 754 k. cal.
Wliere J is the mechanical equivalent of heat and is equal to
Alfyl kg. metre/kilo calorie.
7.3. Adiabatic expansion of a gas. Besides undergoing isothermal
•changes, the gases in an engine e.xpand and compress under such
conditions that no heat enters or leaves the gases. The corresponding
pressure volume changes are said to take place adiabatically and
an adiabatic change is defined as that change iti which a change in
^pressure and volume of a gas take jdace under the condition that no heal is
allowed to enter or have it.
We can understand how an adiabatic change can take place if
we again consider a gas contained inside a cylinder fitted with a
piston. But in this case the outside of the cylinder has been
insulated and the piston is also made of insulating material. If the
piston is depressed, work is done on the gas, and an equivalent
amount of heat is produced. Unlike the case of an isothermal change
however no heat escapes from the gas because the cylinder and piston
are insulated and with the result the temperature of the gas rises. This
is an example of an adiabatic compression. Thus during adiabatic
compression work is done on the gas in compressing it as no heat
leaves the gas during the process, its internal energy »increases by
•an amount equal to the work done on it and consequently rise in
temperature takes place.
If the piston is raised up the gas expands. Now the work is done
by the gas and the equivalent of heat is taken from the gas itself,
which IS therefore cooled. No heat, however enters or leaves the gas
T-K expands, and this is the example of an adiabatic expansion,
thus during adiabatic expansion external work is done by the gas
and as no heat is supplied to the gas from an external source during
■the process, its internal energy decreases by an amount equal to the
work done by the gas and consequently a fall in temperature takes
fire# given above if 710 heat is taken away in the
mil rise if «o heat is supplied in the
the TemnlrVtm^ adiabaUc change
r h ^ ^ ” constant and no heat from outside
lo supplied to the system or taken from it.
13S
ENGINEERING PHYSICS
Hence in an adiabatic change.
(t) Xo heat enters or leaves the gas r.e., H=0
^ [it) But temperature changes as the work is done at the expense
of internal energy.
(n'i) Change in internal energy i.c., increase or decrease in the
internal energy of the gas is exactly equal to the work done on or by
the gas.
From energy gas equation
0=E-\-
ir
j~
we have for a peifect gas.
7.4. Derivation of sdiabatic equation of a grs.
Consider m gms of a perfect gas contained in a perfectly
insulated cylinder fitted with an insulating piston. Let its pressure,
volume and temperature be P,V and T respective!}’. Suppose the
gas is compressed adiabaticallv so that piston moves inwards through
a distance dx. If A is the area of cross-section of the piston the
total force applied is Px ^ and work done by the piston
= Force X distance=P X .4 X df
=Px
=PxdV
where Adx=dV gives tiie change in volume. The heat produced due
to compression causes a rise of temperature c?P and is given by
mC^dl' where Cy is the specific lieat of gas at constant volume.
Hence increase in internal energy dE=‘fnC^dT.
As no heat has been supplied from outside.
We have dH=0
General gas energy equation becomes,
or
dE+~ = Q
mC„ dT^~=0
Now for a perfect gas
PV=RqT
where RQ=inR
Differentiating (ti), we have
PdV+ydP=7nRdT
mR
EXPANSION AND COMPRESSION OF GASES
13&
Substituting the value of dT in (»), we have
PdV+VdP , PdV
mCv
niP
+ --=0
or
^ Pdl^+VdP , PdV ^
Cu ^ J
(C„P(7F+C,JV?P) J-^nPdV=0
or (Cr^+P) PdV -\-C^VdP^o
But
Cp-Cv=^
or
• ■
or
CpJ={R-\-CJ)
CpJ PdV~\-CvJVdP=-0
CpPdV+CvVdP^O
C
Dividing by Ci-PI' and putting ^ =y, we have
C
f'C.PV^CvPV
or
Cp dr dP
-r p
C„ V
dV
+¥=.
Integrating both sides we get,
y loge F*hloge P=constant
logcF^ -|-log€ P=constant
or loge Pr^=constant
or Pr^=constant
It is known as an adiabatic equation of a perfect gas. In other
words if Pi,Vy are the initial and P^y^ the final pressures and volumes
of the gas for an adiabatic change, "then
or
P,
According to the perfect gases equation
AT\ _P2r.
or
T.
P- V
1\ ~ Pi Fi
Substituing for —* in the above relation from the adiabatic gas
HO
ENGINEERING PHYSICS
equation, we get
i.e. TV^ = Constant
Again from the same equation
^2 J
TV
Hence the three
^Constant
forms of an adiabatic gas equation
=Constant
= Constant
are
The
T
,y— 1
= Constant
Now from the equation
7.5. Adiabatic curves are steeper than Isothermal
isothermal equation PP^=sconstant and the adiabatic
Imp>
curves.
equation
V
^constant have been plotted
as shown in the Fig. 7.3. Let
us consider a point P on the^ PV
curves which represents a given
mass of gas under conditions of
pressure, volume and temperature
P, V and T. Now it will be shown
that the slope of the adiabatic
curve is greater than the slope of
the isothermal curve.
The slope of any curve on
the PF diagram is given by
dP
dV'
Fig. 7.3
EXPANSION AND COMPRESSION OF GASES
141.
Case 1 . Isothermal curve I,
For an isothermal curve
PF=constant
PdV + VdP=o
By partial differentiation, we get
Case 2. Adiabatic curve II.
For adiabatic curve
y
PV =constant
yPdV-{-VdP^0
dP P
d\~‘>'V
=yx isothermal slope.
Since y is always greater than unity therefore the slope of the
adiabatic curve is always greater (steeper) than that of the isothermal.
7.6. Work done by a gas during adiabatic expansion. Consider
m gm. of a perfect gas contained in a perfectly insulated cylinder
Let V be the volume P the pressure and T°K the temperature of
the gas. If the piston is allowed to move suddenly outward the
gas expands and does some work. Since no heat is supplied from out-
side, the change is adiabatic and the internal energy is drawn from
the gas due to which the temperature of the gas falls.
If .4 is the area of cross-section of the piston and it moves
through a small distance dx, then the work done for a small chanp-^ in
volume as shown in Fig. 7.2 is given by ®
Work done=dir=Px 4 xdx
Since, the expansion is adiabatic, •
or
dW^K
Bence the total work done when the t*
from the volume F, to the volume F, is given by. ^ ^ adiabatically
142
ENGINEERING PHYSICS
\ _y
Thi^ cxnrosion ^ives tin' work clone by a gas tiuring an adiabatic
exoansion in terms of initial and linal pressures and volumes, ine
same can be expressed in terms of initial and linal temperatures
tlic characteristic gas constant R as given below,
Pi\\^mRa\ and l\V.=‘3nRT.
or P^\\-^P,l\^mR {T^-T,)
mRri\-T ,)
y-1
If one gm. mol. of a gas is considered, then
^_ MR
This shows that the work done an adiabatic ex^
proportional to the fall in temperature increase
an adiabatic compression the temperature of the gas
which is proportional to the work done on the gas.
Examples of adiabatic operations. wonc? no
(i) A sudden compression of air j" temperature
heat leaves the air when compression takes place and P
The condensations and rarefactions in sound waves in air are
examples of perfectly adiabatic operations.
Example. 4. Dry air at normal
Ortc ihird of its original volume under adiabatic cond
the resultant pressure when y=lA.
Initial pressure of the gas Pi=\ atmosphere.
Initial volume Vi=V
(EXPANSION AND COMPRENSION OF GASES
143
Final volume
Final pressure
But
« •
72==— and y=l*4
P77 —constant
■or
or P
1 . V.
)
.=rv(_3.y
3' (3)’' r-3' =(3f =(3)*“
=4-655 Atmospheres
Example 5. (Calculate ih^ rise in temperatun’ when a (ja-s m <'()m 2 Ji\ss»‘d
io 8 Unies its onyinal pressure, assuming that initial tempi ralare is
Initial pressure ~P
Final pressure P^—^P
Initial temperature ri=273 + 27=300®A'
Final temperature Tg— ^
^ve know that
py
— \
:pV
•—constant
or
or
— 1
^py~'^Ti~y
m = (*)
y— 1
1*5
1*5-1
(§-o)=(^) =w
0*5
{T^)
1*5
(300)^*
■- = 8
0*5
or
(300) X 8°*^
( ^ ° =599-8°A:.
[ These calculations have been done on slide rules.]
■ *. Temperature ( 2 = 599*8 — 273=326*8®C
Hence rise in temperature=326*8®—27
=299*8®C
144
ENGINEERING PHYSICS
Example 6. A quauiily oj air at N.T,P. is suddenly compressed
-h-th of its volume. Calculate the temperature attained by air when
y^l'40.
Initial volume say,
Final volume F 2 =-^F
Initial temperature ri=273-f-0
Final temperature T^=% and y = i-40.
For an adiabatic change TF^~^ =constant
r{^r)
y~i _
= ( 5 )’^-* =5
0*4
or log T^ — \ogTx^0-4 log 5
log Ta— log 273 = 0-4 log 5
logr2=0*4 log 5-log 273 = 2-7188
Kence !r2=519'8*^
Rise in temperature=519-8— 273
= 246-8®C
Example 7. A certain qua7itity of air {y=l -4) atN.T.P. f
cally compressed to \th of Us origiruil volume. Find the resuUtngtemp^ fu-
ture and pressure. 1 ' * ‘
Let original volume be equal to F^
^ 1 = 273 "’^^
Pj=76 cms of mercury
Now F 2 =i of its original volume=J Fj.
^ 2 =? and ^3=?
Applying we have
I
Now applying P^Vi
= 273 (i)
= 475 “^'.
=202‘’C.
0.4
= 273X 1.74
, we have
EXPANSION AND COMPRESSION OF GASES
145
=76 {4)‘’‘'
=76 X 6-965
=529*3 cms. of Mercnry
Example 8. One Kg. of air ii at a pressnrp. and temperature of 7
atmosphere and lOO'^C respectively. Find its volume and its internal
energy recohoning internal energy as zero in the' standard state. The air
ts allowed to expand to 4 times its initial volume adiabatically, determine
the pressure , temperature and internal energy. '
According to the gas equation,
PV=RT . j -
P in degree Absolute,
then it IS in Kilo-cal/Kgm/ iC,
X 10^ kg/sq.M. 29*27 K. Cal. /kg/^A^.
7^=273 + 100 =373^K.
Substuting in (t) we have
7X 10^ X 1^=29*27X 3 73
^ 29*27X373
•* — TxTo* — = 0‘156 cubic metre.
Internal energy of 1 kg. of air'=cj„x 1 X change in temperature
= 0*171 X 1 X(373 — 273)
= 0*171 X 1 X 100=17*1 k. cal./kg.
Now for adiabatic expansion PV^ =constant
and
where P^==7 x 10< kg/sq.M,
^1=^1 cubic metre
y=I*4.
^^2— 4^1 cubic metre
7X10"^ (Fi)
1*4
1*4.
or P
{4V^)
7 X 10*
74 10050 kg/sq.M.
4*
As
7n^ we get
PiVx
A
7X10* __ 10050 X 4r,
373 ; jT —
or
^?2 = 373 X
lOOSO
7X 10*
4Fi
=214*2 ^K.
<2 — 214*2 — 273*=— 58*8®c
Internal energy =C„x I X (change in temperature)
—0-171 X 1 (214-2—273)
-0'171X1X(-58-8)
= -12-055 It. cal.
146
ENGINEERING PHYSICS
Example 9. One kg. of a gas expands adiabatically and its fempe-
ralnre is observed to fall from 240°C to 115°C while the volume id
doubled. The gas does 9160 kg. metres of work in the process. Deter-
mine the values of Cp and Ct>.
Work done by the gas 9 1 60 kg. metres
9160 xr*i 1
= Kilo-cal.
426-7
— 21*48 kilo, cal./kg.
Work done during adiabatic operation is numerically equal to
change in internal energy.
Change in internal energy per kg.
= {Tl Tz)
21*48=Cv (2734-240)— (273 + 115)
=0^(513—3 88)
=:0„ 125
= 0-0172
125
Again
where
y=l*4
Op=yO„=l*4X0*0172
= 0-02408
Examole 10. A cylinder containing one-gram molecule
^as^;:r!nLung Iter »atk ar^ am^
until its temperature rose from ZTG to 97 C CalcuMe me wore:
and heat absorbed by the gas (y=lb). Take J-4 2 joules leal.
Work done during adiabatic e.xpansion when the temperature
changes from T^ to is given by the relation
R
W ^ {T^-T,)
y— 1
Where R is the gas constant for one gm- f f lleft
the ratio of the specific heat at constant pressure to the specinc
at constant volume.
7’i=273 +27 = 300®A:
^2 = 273+97 = 370®^'
^ (I-i -T,)
Now
and
W=
Y-l
EXPANSION AND COMPRESSION OF GASES
147
^8*3 (370— 300)_8-3 X70
1-5 — 1 ' 0*5
= 1 1 *62 X 1 0^ joules.
Heat absorbed =
J 4-2
^276-6 cals.
under a pressure of 89 7
pounds per sg. imh ts expanded adiahalically to 39'7 pounds mr so
znch absolute. Find the work done during expZsion. ^ U M I e \
The work done during adiabatic expansion is given by ^ ^
7-1
Also
pyy
constant
or
Now
and
Pi=89-7 pounds per sq. inch
^2=39*7 pounds per sq. inch
y=rl-41
Ti=5 cu. ft.
or
or
Work donA (89-7 X 144 X 5 —
=33000 ft. Ib.
/ ■^:i 44 X 8 ‘ 9 n
i‘4i-rj
ratio
P It. U>.
*■“ 0/5T/ fiabaticaUy through a vo
the work done during expaJiL. and
^ ^p^0‘24s and Ct,^0'/75.
4
14 S
ENGINEERING PHYSICS
Here Pi=
Cp
y=*7=r
28 X 10* Kg./sq. metre
Vi cubic metre
0.245
72 = 5Fi
Cp=0.245
Cv=0.I75
0,175
= 1.4
Y Y
Now for an adiabatic change PiVi =Pa ^'2
1.4 , 1*4
or 28X10 *xT\ =P2(5^i)
28 X 10* — ^ 95X10* kg./sq. metre or technical atmosphere
1.4
Now by applying the gas equation
PxVi P2yt
Ti
vve have
— m * t .
i . , '
> . - '
# «
• V'^
\r -
28X10*Fi 2.95X5
Pi ^^"2
But Pi= 220 + 273 = 493 '’K.
28 X 10* X;r, ^ 2.95XlO* X 5Vj^
493 ^2
2.95X10*
/r 493 X- X 5 —
j ^ 28 X lO*
• t2=259— 273 = — 14X.
Also i?=(Cp-C„) J= (0.245-0.1 75) 420.7
= 29.87 Kg. metre/kg./ K.
jj (fi— r,)
Work done during the expansion per kg.-
29*87 (493 — 259)
1*4 — 1
= 17,470 kg. metre.
E,..,.. ... .< ”«■ • r s
adiabatically expanded ^ ^eaa«re and
o/ specific heats is / ».
pj = 50 atmospheres, J|— 2/4-r
After Expension. „ _,oy ,and y=l 4
p_?, T,=1. and V,-l0y^
EXPANSION AND COMPRESSION OF GASES
H9
Now applying adiabatic expansioni equation, we have
y— I
' •> 1
1*4— 1
0’4,'
- . J J
Temperaturer=140-6— 273s=— 1 32*4®d/ ^
= 140-6®^
Also =
/ p V y , p . 1 ’4
=•'’*(17) (ink)
— 1-4
«=i *99 atmosphers
_ For one gm-mol of the gas, the equation for work done is
given as. uunc i:>
y— 1
2(353 — 140-6)
*“ 1-4 — I
— 1062 cals.
:\
14. A certain volume of dry air at S.T.P. is expanded to
3 times Its volume under (a) isothermal Conditions (6) adiabatic
co^UM^ ^Calculate in each case the final pressure and temperature
The abbreviation S.T.P. . means, standard temperature' and
pressure r.c temperature of 0'*^7=273"5: and pressure 1 atmosphere
or 76 cms. of mercury. Let Fj be the volume of dry air at S.T.P.
(a). Isothermal Conditions.
Initial pressure cms.
Initial volume Vi^Vy
F* = 3F,
FiF,=PJ%
P = _ 76XFj
Fg ^ 3 Fi ‘ =25-33 cms. of Hg
temperature. .
( 0 ) Adiabatic Conditions*
For an adiabatic change.
=P,F,>’
Final volume
Final pressure
9 ♦
or
3Fi./ .
K '
150
ENGINEERING PHYSICS
Pressure P« =
76
Also TiFi’'~‘=T 2 r/~'
Here
ri=273*^
mV^'~^=T^ (3 F,)^''
or
[irT
— 1
= 273
1
273
( 3 )
y— 1
0-4.
= 176 0*if
Temperature=273 — 176
=97^C
7.7. Compressibility of a gas- Consider a gas v<>toe F,
under a pressure P. and subject it to a small increase of pressurej^dP.
The volume will be reduced to V—dV then
volume strain
Change in volume dV
Original volume F
and stress = Applied force/ unit area^^fP
„ Stress dP ^ dP
Hence Elasticity ^-“strain”” dV dC^
is known as Isothermal Elasticity {Euo) and lor aaiaoauc
is knovm as Adiabatic Elasticity {Eadi}-
pressure
K
3 V — =compressibility
t.e., A=-g =- V ^ P
1
Isothermal compressibility=A’,,,„-
and adiabatic compressibility
We can obtain the values of Efso {Ki,o) and Eadi (^a*)
gas laws. The isothermal equation of the gas is
(a) Isothermal conditions. ^ ««
pF=constant
EXPANSION AND COMPRESSION OF GASES
0 ^ .
Differentiating w. r. t. V, we have
«
• •
But
dV
•
• «
p
0
11
But
zr. ^ 1
E»o - P
Ben/ie isothermal compressibility of a gas is
its pressure P. i7 j ^
Adiabatic conditions. The adiabatic
PF^=sConstant.
Differentiating w.r.t. V, we have
' + -^=0
+VVr^ =0
or
^y+Kv =“
or
1
li
t5l
But
.£P
dV
Eadi^yP
Kadi
1
Eadi yP
1
Kadi — ■ y X K^q
Adiabatic compressibility
X Isothermal
compressibility
Also
Kjgo Egtli
Kadia~Eigo ^~'C^ specific heats.
7.8. Clement and Desorme’s Meth,^ fof «.« determination of y(^).
>Otf /
la
152
ENGINEERING PHYSICS
▼OPVM?
1 819 Clement and Desorme’s designed a simple method of measuring
y, the ratio of the two specific heats of a gas.The apparatus consists of
a big, thick walled glass flask placed in a wooden box packed with
saw dust, cotton wool or some
other insulating material to
avoid heat losses. The mouth
of the flask is closed by a
tight-fitting rubber stopper
through a tight tap T from
where connections are taken to
a pump, (ii) a monometer
using light oil or sulphuric acid
and (in) a wide bore stop
(jock S. The flask is filled with
the dry gas at a pre ssure
greater than the atmospheric
pressure P, and the stop cock
S is closed. In order to
completely dry the gas insi(le
the flask some strong sulphuric
acid is kept in the flask. After
some time when the enclosed
gas attains room temperature, the manometer reading is taken. The
pressure Pi is calculated from the equation
Pi=P4-
D
Where P=atraospheric pressure, in cms. ‘liT^the
of the oil, D the density of the mercury and h, the reading
xTel^op cock S is opened quickly for a few rusts
During the ti.me the stop cock ■S cock is closed the
out to equalize the pressure and just whe P completely
gas inside it is at the atmospheric f^ocess is adiabatic,
insulated no heat is either gained or batic conditions the
Thus during the expansion of the gas under a “t jor
temperature falls. The apparatus is then lef m this c
re falls. 1 he apparatus .s ..... adiabatic
some time to allow the enclosed g s 5? n increase
some time to allow tne enciu^cki rauses an increase m
expansion to regain temperature.^^T^
pressure to P 2 , and apart from a g ^which is negligible, the
movement of the liquid in the same. Therefore the
volume of the S-s
increase of pressure from P to P, takes pi surroundings
When the gas acquires the ^Lgrence^A, in the level of
and a steady state has been reached the differe | p the
the liquid in the two limbs is again noted. 1 he pr
enclosed gas is then given by
p
eXPANSiPN AND COMPRESSION OF GASES
... Since the first process is adiabatic, then
133
l-y
or
\T,
}
...{!)
Where Pj is the temperature of the surroundings and T is the
temperature immediately after the adiabatic transformation when the
pressure is atmospheric.
In the second case the temperature again rises from T to Ti and
the pressure rises from P to P, at constant volume. ■
Z=?L
T
or
P 2 A
From (i) and (ti), we have
/.
or
or
fPx
<P2
) (P,
Py
py py
P
A ^(Px
p '.PJ
...(H)
• •
log Pi-log P=xy (log Pi-Iog P,)
Ft-IogP
logPi— logP,
Aliter. Let Pj and be the initial pressure and volume of the
enclosed gas as shown at A in Fig.7.5. If P is the atmospherTc pressure'
which IS the pressure of the gas after the adiabatic exSon and T
the corresponding volume which the xpansion ana K,
th.n the p'oints 1 Ind B «■
PiF/ =PVV
or
P
‘
154
ENGINEERING PHY$IC:S^
shown at C, then the point A and C will be on the same isothermal
curve AC as there is no temperature change from A to G.
Fig. 7.5
Pg Pi
vr p«
Substituling the value of
PJ ~P
^ in (t) we have
(«*)
or y(IogPx— logP 2 )=logPi— logP
log Pi— log P ...(m)
log Pi— log P 2
This relation is the same as desired above.
Thus y can be calulated if P, Pi and P^ are known.
A very convenient approximation can be made if
onlv sliehtlv CTeater than P. an adjustment easily
a^ging p?esfure P, in flask to be initially slightly greater than the
atmospharic pressure.
Rearranging and substituting the values in relation (m),
hiP
log P+
log
Pi
P
D
log(
log
P+hiP
D
P+
htP
D
log TTZ
P
EXPRESSION AND COMPRESSION OF GASES
15S
But \oge — ..from a well known formula in
mathematics where x<l.
V
D
D
1
D
htP
D '^ D
h.
p
(A| — A,)
fA
If A„ are small compared with P. Since is then equat
to P, to very good approximation, we have finally
y^^lL
(Ai-A,}/P ~
This method gives fairly accurate value of v and is venr
convenient to use in the laboratory. va e oi y ana is verjr
"?<=«?sary to make the prerssure atmospheric. The
fnd a^r trim’’® gas inside therefore becomes less than the atmospheric
«et UD andTt t ® "'^y oscillations are
ataoLher;;! considerable time for the pressure to become
atmospheric. The stop cock S should be closed at the instant when
diSt to ludge^“°"’ atmospheric, which is very
Expected Questions
I. (a) Explain and state the general gas equation.
lUustiSe S Isothermal change (<i) an adiabatic^ change.
a perfect gas und^gorng^an adulSic c^^ volume and temp^^re
many «q adiabatil^I^y Z^"a)
5. Explain with a neat sketch the method of finding out the value of
CHAPTER VIIT
KINETIC THEORY OF GASES
8*1. Introduction. Danial Bernouli in 1730 suggested akinetic
■theory of matter based on two assumpcibns.
{i) that matter is not continuous but consists of small
aggregates, called molecules with inter spaces between them.
(ii) that the molecules are in a continuous state of agitation.
Both these assumptions are supported by experimental evidence.
The kinetic theory of solids and liquids is not yet in a .
development but the kinetic theory of gases can fully account for
various phenomenon in gases.
8.2. Kinetic Theory of Gases. The kinetic theory of gases has been
highly succesful in e.'cplaining the behaviour of gases. The essential
feati^es made of very tiny particles called molecules.
The molecules of a gas are alike and differ from the molecules of other
The size of the molecules is negligible i.e ■nolecules
are riet poi^t masses as compared to the distance
iiii) The molecules are . continuously m moUon. Their motion
is absolutely haphazard and ^The eas is said to be in a
SrS? ^ho" increases
""^rTritead, st^ ^,o.cules are continous. *
against each other and with the walls
of the containing vessel. Between two
collisions a molecule moves m a
straight line. The paths they follow
while moving inside the gas are
irregular and zig-zag as shown. c
individual lengths of these paths vary
winthin wide limits but the averse
value of a large number of such paths
lias a value at any temperature known
as mean free path.
156
KINETIC THEORY OF GASES 157
Mean free path (A)= ...+dn
^ A ' 4 ^ «5
etc. are the individual paths.
(w) The time during which a collision takes place is negligible as
compared to the time required to traverse the free path i.e., the
collisions are instantaneous.
(vi) The molecules in their motion collide with one another and
with the walls of the containing vessel. At each collision their
velocities are altered in direction and magnitude, yet in the steady
state the collisions do not affect the molecular density of the gas i.e.,
the number of molecules per c.c. of^he gas remains constant.
/
(m) The molecules are perfect elastic spheres and exert negligible
force of attraction or repulsion on another or on the walls of the
containing vessel.^ In other words their encounters (collisions) with
each other an_dj^fh the wall s of co ntainigg vessel are perfectly elastic
and no energy>is lost during encounter or collision.
We have seen tb^t thnfiOTecules in
directions with all possible velocities,
types of velocities as given below.
a gas move randomly in all
We normally consider three
velocity or average velocity. If n molecules of a gas having
velocities are considered, then the mean or average velocity
n
(w) Root mean square (R.M.S.) velocity. Root mean square velocity
of the n molecules is the square root of the mean of the squares
of the individual velocities. Mathematically,
Cr. ,n. +rn^
V ^
■ It is found that although due to the repeated collisions the
individual velocities may change, but the r. m. «. velocity remains
constant so long as temperature remains constant.
relaW velocity can also be' calculated from the
Cr.
m* g
I 3TK
Af m
Where !r=absolute temperature of the gas.
.ff=Boltzmann’s constant and
m~mass per molecule of the gas.
Cf m. a cc
158
ENGINEERING PHYSICS
{in) Most probable velocity. If the molecules of a gas, possess-
ing different individual velocities be grouped together, then it
it is seen that the number of particles possessing different
velocities is different. If the distribution of the velocities of
molecules is plotted as function of the number of molecules.
Fig. 8.2
obtain a curve known as numb“
shown in Fig.8.2. From the curve it is found that
of rTlecules^ at any temperature move with a ,f,“ies?nd
velocity is known as most probable velocity of the m es
maximim number of molecules (as shown by ordinate (a 6) have t
most probable velocity.
Most probable velocity is defined as
largest number of molecules in a gas are found to move at y 9
The value of this velocity is given by the formula as
IzkT'
Vm— /
V m
here k, T and m have the usual meanings.
(f„) Relation between mean velocity and
Clerk Maxwell has shown how the J equilibrium and from
rh1^tory"tL re\S between" o’ and C has been deduced out. They
are related as follows. ^
.-J^c
This numerical factor ^ J _ is very nearly equal to
_12
13
SO that c
is about
12
of C the mean velocity being some what smaller than the
13
root mean square velocity.
KINETIC THEORY OF GASES
159
8.3. Pressure of a gas. The molecules in their *•
will not only collide with one another, but also impinge on^he
he "" sfnrtf' <-> definite momen um to
thewa s._ Since the number of molecules making such imnarfc
the walls IS very great, the walls will experienced almost dnt inn ’
force per unit area known as pressure. Hence acrordid . ,
theory, the pressure exerted by a gas is merely the totaf Emetic
communicated to the walls of the Lntainine vessel momentum
unit area due to the impactsdf ?he molecules of th*“°"'^
expression for the pressure exerted by a gas can be found as f?li;ws
side i™vft^its^dgdtadllfl^^ t" ^ of
infVeti aTe%drtly*'’e"‘lI^:ic“‘'nrd:re'' 1"/ ‘he"co"tain-
vessel, each of mass m '’o n molecules in the
— ^ 8— 2 , « * cube. Then
The molecule striked thlfde'!f^^th" P®7®"dicular to the face A
thenit^^iU(i*:;)/‘~‘h^^ oollision U
before and Iff '°"l?*on. Similarly it* collision
>- is
Th» . =
travels b™k agafa to°I ^ A' It st k
160
ENGINEERING PHYSICS
of the sides of the cube). The time taken by the molecule to go to
the other face and then to come back is
u
u
« •
Numer of impacts per second= “
U\
The rate of change of momentum= 2 mWi . ^
mwi
I
Now according to Newton's second law, the rate of change of
momentum is equal to the impressed force. If is the impressed
force, then
F.=
mu
2
I
Similarly the force F^ due to the impact of another molecule
having a velocity C^. whose three rectangular components are
and 14>2. is given by
mu
z
and o on. i J n ^ t + 1 ,,^
Hence the total force! jPx onThe face A due to impacts of all the
„ molecules is given by V .
— wr
‘=-y-
Since pressure is the force-per unit afea, the pressure P. on the
. • ‘ '•
face A is given by
p ^ =
m
V
Fx
Area; :lx{lxl)
(ui* +^ 2 ^ + • • • +
t*
<rvv
. V S'^ to the y axis the pressure.
Similarly for the face B perpendicular to the r axi
Py=T
" and for the face C perpendicular to the ^-axis, the pressure
P^=y («Ji2+u,2^ + ...+«Jn*)
Since tL pressure exerted by a gas is the same in all directions,
the a"^er:ge%r^essure P of the gas is gwen by
Px-hPy±^
P=
3
KINETIC THEORY OF GASES
161
'VU.
X'>
V) f«*
\
TV'- '-*-1—,*-^ «=*- .
[(’^ 1* + * • • + (^1* +*’ 2 ® • • • +t»n*J
^ [(«r+n^+«-i^)(w2*+v2=*H-w2“)
3 ^ -f ••♦+(«^n*4-t'n^+M^n*)
•=-Li|+v> /
I m
C V
= Uv
-Vv
^ AWt ^ 3
1 mnC-
Q c ^'-. ^ =
(Ci=+C^ + ...+C„"-)
p = A __
Where"
and C is known as the root mean square velocity of tlie molecules.
If M be the total mass of the gas tlien,
3 / =mn
and
M
Y-»
where p is the density, then
P=--l PC-
Relation between pressure and kinetic energy. The pressure exerted by
the gas is given by the relation,
-P=i pc- (as proved above)
-=!xipe^
= f £
where E is the mean kinetic energy per unit volume of the gas.
Hence the pressure of the gas is equal to § of the translational
kinetic energy of the molecules in a unit volume.
Kinetic energy per gm. -molecule. From the relation,
P=
1 mwC*
we have
Now according to the perfect gas equation
PV^RT
^“Volume per gm. molecule
Universal gas constant
If in relation (i) y is also the gram molecular volume, then
mn^M
where
ENGINEERING PHYS’CS
will be the mass per gm-molecule. Hence the relation (t) becomes
or i -1/C2=| RT
Where ^ is the kinetic energy per gm. molecule of any gas
at the absolute temperature T.
Hence the kinetic e7icrgi/ per gm-molecule of any gas at the absolute
temperature T is equal to | RT.
Root-mean spare velocity. From the relation, it is clear
that
-J
iP
Hence this expression gives tlie root mean square velocity in terms
of the pressure of a gas P and the density of the gas p.
M
Further we have p = '^where M is the molecular weight and V is
the molar volume.
M
or
But PV=RT (gas equation)
or
C- i
f^RT
V
M
But
M=^inN
where A=Avogadro's number.
•
C= A
fsRT
• •
\I rnN
Now ^=/:.
A
•
• •
C=
j 3kT
...{Hi}
<
II
1 m
Since
hence from equation (n) we have
. (tr)
V 3rT
Whp*€ r is the characteristic gas constant.
«-4. Kinetic interpretation of Temperatnre. We have derived the
relation mnC'^ for the pressure of a gas, where n is the number of
KINETiC THEORY OF GASES
163
molecules in a unit volume of the gas. Multiplying by V, the volume of
a gram molecule of tlie gas, we have,
PF =imnVC-^=imNC^
Where the number of molecules in one gram molecule of
the gas, or the Avogadro number as it is generally called.
Let M=mN
PK=P/C*
According to the perfect gas equation.
pv=Jir
Equating (t) and (ii) we have
In other words. is proportional to the absolute
At r=o, c — 0
temperature T.
Hence the absolute zero of temperature is that temperature at
which the linear velocities of the molecules of a gas are reduced
to zero.
Now
M mN
or
or
4t
Where RjN is a constant for
constant denoted by the letter k.
a gas and is called Boltzmann's
So that.
1 mcr^ =_i
Where ^ mC* is the mean kinetic energy of a molecide of the gas.
Thus mean kinetic energy of a molecule of the gas,
and K. E. of a gram molecule of the
JcT.
k.N.T,
N
In other words, the mean
gram molecule in a given mass of
temperature T.
K. E. per molecule or K. E, of the
a gas IS proportional to the absolute
engineering physics
This fact is referred as Kinetic interpretation of temperature.
8 5. Deduction of simple gas Laws. The various gas laws mav be
deduced cn the basis of the kinetic theory as follows :
(») Bojie’s Law. We know that the pressure of a gas
IS given by tlie relation, P^^ivirC- where n is tl.e number of molecules
in a unit volume of the gas.
Multiplying both sides by V, the volume of tlie gas, we have
PV=^7nnVC^ mNC^
W hero =7d , the total number of molecules in volume V of
tlie gas.
a constant, because is constant
Hence Pr=Coustant.
Tlie heat energy of a body is the energy of methn of its n-ojeciiles.
WTien the temperature of the gas is kept "constant the heat energy
of a given mass and hence total kinetic energy cf its molecules is
constant.
Total K.E.—^ mnC^ is constant at constant temperature.
It follows, therefore that at constant teinptrature, Pr=Constant.
Boyle's Law is thus easily accounted for.
(if) Perfect gas equation. We have the relation, P=^mnC^ =^PC^
wliere ?n«=density of the gas.
Multiplying by V, the volume of a gram molecule of the gas, w’e
therefore have.
PV=^
WHiere Pr— T/'=the mass of a gram molecule of the gas, and
therefore \ il/C- its kinetic energy.
Now, we know, that the absolute temperature of a gas is propor-
tional to its kinetic energy so that,
PFoc I MC^o: T
or PV=TX 3 constant or PV=PT the standard gas equation
where R is tlie gas constant for given molecule of the gas.
{Hi) Charles’ Law- It has been proved above that PV=RT.
If the pressure is kept constant — ^ is constant i.e ,VccT . In other
words, the volume of a gas is directly proportional to the absolute
temperature, which is Charie’s law.
(aV) Avogadro's Hypothesis. Suppose there are two gases and
and Cl represent the mass per molecule, number of molecules per c.c.
and root mean square velocity repectively for one gas and and
KINETIC THEORY OF GASES
165
the correspDnding values for the other gas. If the two gases exert
the same pressure P, then
If the two gases are also at the same temperature there will be no
trans^r of heat energy from one to the other when tliey are mixed
^P* This IS possible only if the mean kinetic energy per molecule in
the two gases is equal. In other words.
A rW c. ^
X. i ^ ^ f I j
From (») and (U) we have
In other words, equal volumes of two gases, unier the same conditions
oj pressure and temperature , contain the same number of molecules, and
ms IS Avogadro s Hypothesis
For convenience, .V (tt K) is usually taken to be the number of
Number^^ ^ molecule of the gas and is called Avogadro 's
e i - Calcutafe th^ root mean square velocity of the tmlecules
ofyurojenatOC. The density of nitrogen at N.T.P. is P25 am.!
hire. Density of mercury 13-59 gmfc.c. aiul g =981 cml-secK
{A. 3f. 1. E.)
Normal pressure P=76-00 cm of mercury column.
= 76X 13-59X981 dynes/cm®.
1-25
density
P =
1000
=0-00125 gm/c. C.
P- M. S. velocity, C=
=3 / 3 X 76X 13-69 X981
V ~ 0 00125
=4-93 X 10* cm/sec.
molecule of oxygen al 27°G. clnverltl^ inlo KmUhowT “
The root mean squre velocity ^
ENGINEERING PHYSICS
Now R=Nk=&02 X 10^3 X 1*38 X 10-®=8*31 X 10^ ergs/degree =8*31
joules/degree
Where ^=Avogadro’s number
/;=Boltzmann's constant.
itf=MoIecular weight of oxygen=2X 16=32
273+27 = 300®ir.
(y _ / 3X8 3I X10^X300
32
=48,350 cm/sec.
=483*50 metres/sec.
Converting it into km/hour, we get.
^ 483*5X 3600
Example 3. Given that the gram-molecular weight of a gaa occupies
22-4 litres at N.T.P., calculate the r.m.s. velocities of Hydrogen, gas,
(/) at O^C and {ii) at 50^C. Assume the usual values for any physical
constant. Density of mercury^lS'6 gmsjc.c. and g=981 cm.fsec^.
{A. M. r. B.)
(i) Molecular wt. of Hydrogen in gms
= 2 .
2
Density of gas at 0^67,= gm./c.c.
^ ^ 22400 ® '
R.M.S. ve
where
P=76 XI 3*6X981 dynes/sq. cm.
r!_ /3X 76X13-6X981X22400
V 2
= 146400 cms./sec.
(t*) Assuming that pressure remains constant,
R.M.S. Velocity cC's/T
R.M.S. Vel. at S0°C I'T _ / 273+^
R.M.S. Vel. at 0"C "V V 273 + 0
/ 323
R.M.S. Vel. at SO'^C'^ 146400 X J
= 159200 cms./sec.
ai592'00 metres/sec.
KINETIC THEORY OF GASES
167
Example 4- 32 grama of oxygen at N.T.P. occupy 22'3 litres. Deduce
the r.m.s. velocity of the molecules at 20^C. {A.M.I.E.)
22*3X103“
Pq=276X 1 3*6 X 98 1 dynes/sq. cm.
a
-j’-t-J
3 X 76 X 13*6X981 X 22300
32
n 2
Now
=45040 cm. /sec.
273 273
n 2
^20
273+^ 273 t 20
or
Or
^ 20 = ^0 ^
273
293
273
=45040X
273
=47680 cms./sec
j
Example 5. Find the temperature at which r.m.s. velocity of Hydroaen
Will be double of its value at N.T.P. when the jnessure remains constant.
R.M.S. Velocity at
or
2
1
Squaring both sides
4
1 ^
or
f=:
j
j
R.M.S. Velocity at PC f p
273
J
im+t
273
273+/
273
273
«=273X4— 273 = 819®C
a< 29“c if of 1 gm. of Helium
at L tfS— 8 31x10^ ergsrK and molecular wt. of helium is 4.
K.E. per molecuIe=~ JcT.
where h is the Boltzmann's constant.
4 then^ Avogadoir’s number and molecular weight of helium is
No. of molecules in one gm.=
N
168
ENGINEERING PHYSICS
*. ■A'.J?. per gm. of Helium = ~ X— i-r
4 2
But
N
K.E. per gm. of Heliuni=— X — x— T
4 2 iV
= i RT
=iX8*3IXlO’X300
^9-1 X 10® ergs
= 910 joules
Example 7. The jnass of a)i oxygen molecule is 5'28xl0-^^ gm. and its
mean velocity at X.T.P. is 42500 cm.jsec. Calculate the mean kinetic
energy of an oxygen molecule at 50°C.
We know
Squaring
or
C=
V 8
Root mean square velocity of oxygen at O^C.
1 X42500
=46100 cms./sec.
Also
Cac\/T
or
Qo'-^273+/_273+50
Ca2 273
273
or
2_323
<•50 -273
X{46100)2
= 2*52X 10® cm./sec.
The mean kinetic energy at 50®(7.
=J m Cq^
= JX5-28 X 10-23 X2-52X 10®
= 6*65 X I0-“ ergs
= 6-65 X 10-21 joules
Example 8. Given Avogadro’s nwnher N as 6'02xJ0‘^^ and Boltzmann s
constant-l -37x40-^ ergs per^C. Calculate (a) The kinetic energy oj
translation of an oxygen inolecule at 27°C [h) The total kirietxc energy J
an oxygen molecule at 27°C. (c) The total kinetic energy in joules oj
gram molecule of oxygen at 27^C,
KINETIC THEORY OF GASES
169
Oxygen is a diatomic molecule having three degrees of freedom of
the motion of translation and two degrees of freedom of the motion
of rotation i.e.. in all five degrees of freedom.
[а] K.E. of translation per molecule
= — kT.
2
X 1-37 X I 0-«X (273 + 27)
= 6- 1 65 X 10-^* ergs
= 6165 X 10 “* joules
(б) Total kinetic energy per molecule
= — kT.
==Y ^ (273 + 27)
= 10-275 X 10“^* ergs
= 1 0*275 X 10"-^ joules
(c) Total K.E. per gm. molecule
=4 kTxX
= 10-275 X lO-’^X 6 02X 10-®
= 6185-55 joules
Example 9. Cahulaie the specific heats of Helium at constant pressure
constant volume, giren that the iUtisity of Helium at K.T.P. is 0’28
gm.jhtre and J =4-2x10'’ crgsjcal.
As the density of the gas is 0*18 gm./Utre
• •• Volume of the
We know from the
gas for I gm.=^ C.C.
gas equation, PV=RT
or
(76X13-6X981) X
273
R if 22 is in heat uni ts.
1000
0-18
Now
170
ENGINEERING PHYSICS
If R is not in heat units, then divide R by J.
Cp=±RIJ
_ 5 X 76 X 13-6X981 XIOOO
2X0 18X273 X4-2X 10^
= 1-23
^lod C'i,=-ixi-23
*=0-74
8 6. Degrees of freedom. To completely describe the motion of
a particle in one plane only two quantities must be known say its
two rectan^lar co-ordinates. Similarly for a particle moving in
space three independent co-ordinates mu;t be known to describe its
motion. A molecule in a rigid body can have three motions of
vibration along any of the three co-ordinate axes in addition to
its three motions of translation. Hence to completely describe the
state of motion of a rigid body six co-ordinates are required.
The degrees of freedom may, therefore be defined as the total number of
co-ordinates required to specify completely the position and configuration
of any dynamical system.
Take the case of a monoatomic gas molecule i.e., a molecule
made up of single atom only. This molecule can rotate about its
Fig. 8.4.
polar axis and can also move bodily in three mutually perpendicular
planes a:, r and Z as shown in Fig. 8.4. J^^nt/nonal
lational degrees of freedom and one rotational, but the
degree of freedom has such a low moment of inertia that the effect
of this is negligible. Hence a molecule of monoatomic gas Has 3
degrees of freedom of translation only.
The molecule of a diatomic gas such as “yff "
nitrogen can be considered to have the shape of a *■ f j^^ia
inFi|. 8.5. Any such molecule has apprec.able
and it is capable of rotational as well as ? the end
a molecule can have a polar rotational motion as
KINETIC THEORY OF GASES
17L
view and it may also have two perpendicular rotational movements
as shown in the elevation and plan. But the moment of inertia of
fl'Wi t CUL £ mol Bcaie
\
£ L I' \/ A TtOM
SA/O V/£W
Fig. as.
axis i.c.. about the axis joining
this direction a Th perpendicular tc
pnerally neglected. Thus we have three degrees of freedom of
ThTst SlnTFr",*,""! rotational^to give five ^ all
har4 4en resolJj' ' . ’ "-hich the rotational moment
have been resolved into two moments about two axes il and 21
perpendicular to the line joining the molecule. ' '
more’Itoms‘^’‘*nTh' as O,. CO,. H,0 consists of 3 or
S can be arrln a*" triatomic molecule
harabDrec1ahr,^fl f Any such molecule
afwXTfranL^?nTf it ia . also capable of rotational
into three motinrxi rotational motion can be resolved
as shX in the Fig n "
degrees of freedom^.e.; 3<tknsTa«on"af:L''d‘3To'taS'* "
172
ENGINEERING PHYSCIS
In general, the more complex the molecule, the greater will be
the number of degrees of freedom possessed by it.
molecule
MOLf.CULE
MOteCULE
(b)
Fig. 8.6.
Law of equipartition of Energy. This law states, that “the
total kinetic energy of a molecule in each degree of freedom is ^ hTM
This law was first formulated by Maxwell in 1 859 for translational
degrees of freedom but later on Boltzmann extended it further to the
rotational degrees of freedom etc.
In the treatment of kinetic theory of gases it is assumed that
in the steady state
n, V and u’ are the average values of the component velocities.
^ 7nn- = ^ 7nr- = J ^ jn C-
But ^ tnC^=^l kT
The K.E. per molecule per degree of freedom
. I kT=\kT.
Thus we find that the total energy is divided equally among all
the three degrees of freedom. The result is true for any number of
degrees of freedom and is known as the law of equipartition of energy.
It states that for any rhjnamical system in thermal equilibrium the total
energy is divided equally among all the degrees of freedom and the. energy
■associated with one molecule per degree of freedom is equal to j kT.
The K E. per gram molecule per drgree of freedom
=l kTxy^l RT
where N is the Avogadro’s number and R the gas constant for a gram
molecule.
8.8. Ratio of Specific Heats.
H) Monoatmonic gas molecules. A mono atomic gas mojccule
has only three degrees of freedom as it has only a motion of trans-
lation in space,
.-. Total per gm. molecule T°K=^'<\RT.
— j?r.
2
kinetic theory cf gases
173
and total K.E. per gm. molecule at (r+ 1)
But the increase in the total K.E. per gm. molecule per degree
rise of temperature is called the molar specific heat at constant
volume i.c., Cy
C„=if(7’+ 1 ) -- J?.
But
Cf, — Cij=R
Cp=C*t,-h72^4 R-\-R=4li^
Cp 2 5
^ =1.66
Lr ^
'y
(tt) Diatomic gas molecules. As already explained the molecule
of a diatomic gas has 5 degrees of freedom.
Total K.E. per gm. molecule at Rl'
and total K.E. per gm. molecule at (2'-|- i)®A'=-— 1)
Increase in total energy per gm. molecule per degree rise of
temperature, Ct,=-^A(r+l) — ~R.
But
Cp — C^-^-R^—^ R-^J(~—^R.
Hence
,(*■“■) Triatomic gas molecules. As already explained the molecule of a
triatomic gas has 6 degrees of freedom.
Total per gm, molecule T°Kss.6X^RT
=^^RT.
and total K.E. per gm. molecule at {T f I)°A:=3A(T-|-I).
Cp^molar specihe heat=3i2(3'-f i)—
Cp^ZR^R=4R
greate”tSLru:^ty « always
174
ENGINEERING PHYSICS
Kinetic energy and ratio of specific heats of gas molecules
Molecule
T'egrees of freedom
Average K E
Y
Translate
Rotation ,
Total
1
Munoatcmic
1 3
1
1 .
0
3
.±KT
1-66
Diatomic
1
3
2
5
-^KT
1 4
Polyatomic [tri-atomic]
3
3
6
~KT
1-33
8.9. Explanation of three states of matter on the basis of kinetc theory.
Every substance whether solid, liquid or gaseous is made up of many
millions of elementary particles known as molecules of the substance.
John Dalton in 1803 was the first person to make use of this conception
of matter. Later it was realised that molecules of a substance had
some form of movement and there developed the kinetic theory of
Matter, which in the hands of Clark Maxwell enabled the properties
of a gas to be explained.
According to the kinetic theory, the molecules of a gp move
about in all directions, making collisions for a short time with each
other and with the walls of the containing vessel and rebounding
from them. The motion of the molecules is a random or haphazard
one and the average of the impacts per second of the molecules on
one side of the container creates certain pressure there, which is the
pressure of the gas.
Particles of matter have an attration for each other which decreases
as their distance apart increases.
1 SOLIDS, (t) Shape and Volume. The molecules of a solid
are very SVy palLd. f hey attract each other with a very large
torcedue to cohesion. They can not move about
simply vibrate about their mean positions. This is why a solid
definite shape and volume.
(ii) Elxulicity and Rigidity. "=‘‘^’'"?L^Xance’beJween
change the size or shape of a solid it changes ^ this force
the molecules or their relative 9” and the
the molecules resume their original position a g , j the
solid regains its original volume or shape. This explain
property of elasticity and rigidity of the solids. :„^„ases
[Ui) Expansion. On heating the velocity ««
They Vibrate with greater and greater amplitude and
increases. The solids, therefore, e.xpand on heating.
(iv) Melting. On further Seating ^ [“ul/caiTleave its
amplitude of vibrations becomes so large that a ^
mean position The solid then changes into liquid form
^INETJC THEORY OF GASES
175
temperature is known as mclling point. The heat now supplied to the
solid does not further increase the velocity of the molecales but is us -d
in breaking the rigid structure of the molecules and takes them
further away against their mutual attraction. This is then tl'.e latent
heat of fusion.
2. Liquids. (t) Shupn and Volume. The molecules of a licjuid
are fartl'er apart than those of a solid. They attract eacli other with
smaller force of cohesion. They can not leave the liquid but are free
to move about anywhere in it. This is v/hy a liquid has a definite
volume but no shape of its own.
(ii) Surface Tension. A molecule lying within the liquid
is attracted equally ia all directions and hence the resultant
force on it is zero. The molecules lying on the surface are attracted
side ways and downward only. This explains why the free surface
of a liquid behaves like a 'stretched membrane and gives rise to
surface tension.
{Hi) Cooling due to Evaporation. Different molecules have different
velocities and the temperature of a liquid is determined by the mean
square velocity. The molecules moving towards the surface are pulled
downwards by the molecules lying b*low and so can not leave the
liquid surface. But some of them which possess velocities much
greater than, the average, manage to escape. This explains
Evaporation.
As the molecules possessing large velocities escaps, the mean
square velocity of the molecules left behind decreases and, therefore,
the temperature falls. This is why evaporation causes cooling.
(tv) Diffusion. The liquids slowly diffuse in to each other due to
"the motion of the molecules.
(v) Expansion. On heating the velocity of the liquid molecules
also increases and they travel greater and greater distances The
iiquid thus expands on heating.
{vi) Boiling. On further heating a stage reaches when the velocity
of molecules becomes so large that they can overcome the force of
attraction of other molecules and escape form the surface The liquid
begins to boil and the temperature is called boiling point The
heat now supplied enables the molecules to over come the force of
attraction and the temprature does not rise further. This is called
latent heat of vaporisation. canea
werv The molecules of a gas a
very much farther apart than those of a solid or a liquid This c
stelm" changes into 1700 c.c.
sp“ e avaflable^to Th' 'cee to wander over the fnti
Zpe onJs ow„'° "«‘her a volume noi
ENGINEERING PHYSICS
lie
{ii) Boyles law. the gas is contained in a vessel, its
molecules continuously strike against the walls and are turned back
with equal and opposine velocities. The rate of change of mountum
per unit area gives the pressure exerted by gas. When the volume
of the vessel is increased, the number of molecules striking a unit area
of walls in one second decreases. Hence tlie pressure is inversely
preportional to volume. This explains Boyle’s law.
{Hi) RegnaulVs Law. When the vessel is heated the temprature
of the gas and therefore, the velocity of the molccuics increases. This
brings about an increase ip. the rate of change of momentum per unit
area p^-r The pressure of a gas, therefore, is proprotional to the
absolute te-fnperaturc.
Mean free path. According to kinetic theory of gases it is
assumed that the molecules of a gas are in a state of continuous motion
and their motion is absolutely haphazard and irregular. During their
movement they go on colliding against each other. Between two
collisions tlie molecules travel a certain distance in a straight line. This
distance is called the free path. The particular direction of motion
and velocity is changed only during collisions with other molecules
or with the walls of the contiining vessel. The molecule travels
undisturbed in a straight line till it
approaches very close another
iiulecule when it is suddenly deflected
due to the influence of the latter such
a meeting of molecules is known as
encounter and is of very short duration
and is usually accompanied by changes
in tiie velocities of the molecules.
Hence the path of a single molecule
will consist of a series of short zig
zag paths as shown in Fig. 8.7. Some
of these paths will be long, otiiers will
be short but the average of a large
number of such paths has a definite value at any temperature and
is known as mean free path (A).
If di, d^, dg, etc. are the individual paths then the, mean free
path is given by,
diH-dg+dsH dfi D
~N ~ N.
Where D is the total path travelled in N collisions and is given by
X)=diH-d2d-t^3H
Hence the mean free path A is defined as the average distance travell-
ed hy a molecule between two successive collisions.
D
GAS MOLECULES
o o
Q
Fig. S.7
A =
NgTiC THE DRY OF GASES
177
8.11. Calculation of the mean free path. To simplify the
calculations we suppose that,
(1) Only the molecule under consideration is in motion and all
other molecules are at rest.
(2) The molecule has a sphere of inihience around it. The
radius of this sphere of influence is equal to tlie diameter d of the
molecule as shown in the Fig. 8.8 (a)
Sphere of influence
(a)
Justcollided Collided
(c)
Fig. 8.8.
iheones
those mo^leru'iesThlceL^rrsXw d'T
collision is shown in the f4 A
rest. Asitmoves iL snherp^^^ molecules are at
a cylinder whose cross-sectional area~Tr"J"^^^H second
distance traversed in one second ieC ^®"Sth is the
uqder consideration collides with all ’ , the molecule
Which lie .vith a cyirnd^ a„"d iTn^tht^
••• the volume of the cylind^T^^aC.
Let the number of molecules per c. c. be n, then
Nymber of
collisions per sec.
molecules
Fig. 8.9
in the cylinder=,i2o.,_the number of
17S
ENGINEERING PHYSICS
Average time between two successVe collisions=^^^ 2 ^ sec
Hence average distance between two successive collisions
1
Tzd'^Cn
xC =
1
Tcd^n
Mean free path,
Tf m is the mass of one molecule, then
mn=p, the density of the gas.
• •
A=r 4" =
m
•rdhjin ■7:0} 9
Hence the mean free path A is inversely proportional to the
density of the gas P which itself varies directly as the pressure and
inversely as tlie absolute temperature. Therefore the mean P,.
irinvers^ely proportional to pressure and directly proportional to the
absolute temperature.
Example 10. The mean ?. free path of the molecules of « »
2 xl 0 -^m mhen there are T25xl0'^ molecules per cm». Compute
the diameter of the nwlecules.
Here
Diametar
71=1-25 XIO”
A= 2 X 10'® cms.
d= 1
Using the formula,
rf2 =
]
or
d=
7:7 iA
1
V ”nA
All the letters are having the usual meanings and putting the given
values we find,
1
d=
V TCX 1-25 X lO^’X^ 10 ®
I
10-'
28 03
■y/ 71X2-5X10^*
= 3.5Xl0-®cm.
KINETi: THEORY OF GASES
179
compared to the thermal velocity of the molecules
in the direction OX. Tlien, tiie
velocity of the layer in contact
with the surface is zero and it
increases as we go along OY
in a direction perpendicular to
OX at a uniform rate
dy'
Consider a layer BB at a
certain distance from the fixed
surface OX. The velocity with
which the gas flows in this
layer is v. Consider two layers
AA and C C above and below B B
^ ® , respectively at a distance A equal
molecules moving
ween the t“wo°ayerT °‘ '"Hisions while moving bet!
The velocity of gas in the layer A ^=v+A —
dy
dv
dy
and the velocity of gas in the layer C C=v~X
‘hejmal velocities the molecule are moving in all
that on an average one sixth of thf 'Zi f ^ respectively, so
o»e ojois in one particXr direct Th" ®® • parallel to any
interchanp of morec^^bJtwtTlriaye^s^,^'^. d
then the number of -"olecules’plstTdo—
unit area of the layer B B in one second -“i.
P- ""it "^ea^per second by the layer
Similarly the number of molecules moving upwards per unit area
■of the layer BB in one second
^:iz:k:zs “•* - — -
Net momentum lost by the lflt»<^r a a
^ AA per unit area per sec.
mnc
6
!('««)-(-* I)}
ENGINEERING PHYSICS
=J mncA
dv
dy
The layer CC below BB gains the same amount of momentum. -
Hence tlie layer .4^ above BB tends to accelerate its motion and
the layer CC below BB tends to retard its motion.
The backward dragging force per unit area -gain or loss of
momentum per unit area in one second.
B=\')nncX
dv
ly'
dv
This must be equal to the tangential force 7-^ acting per unit
unit area of the layer BB due to viscosity, V being the co-efficient of
viscosity of the gas.
dv
dy
or
=iPcA
[./ mn=p]
nressure but A decreases in
The density of a gas increases with pressu
the same ratio so that pA remains constant.
Hence v is independent of pressure.
i„crels:s
tional to -v/^^where T is the absolute temperature.
VccV'^ -
1
Vo
V
Expected Questions
I (o) state the assumptions of Kinetic theory of gasts and
lead to gas laws of a perfect gas.
(6) Distinguish between the following ... '
,i) Mean or average velocity («) root mean square
probaWe velocity of an assemblage of molecule .
2. (a) Derive an expression for the pressure of a gas on t
tlieory. i , .. translational Hnetic
(6) Prove that the pressure of a gas is equal to y , (AMJ-S.)
energy of the molecules in a unit volume.
KINH'IC THEORY OF GASES
181
3. (a) Outline the essential features of kinetic theory of gases and apply
it to account for the simple gas laws.
(6) What is the interpretation of temperature on this theory ?
4. Deduce the values of two specific heats and of their ratio y for (a)
monoatomic gases (6) diatomic gases (c) Poly-atomic gases on the basis of
the principle of equipartition of energy. {A.M.I.E.)
5- What is the meaning of mean free path of the molecules of a gas ?
Show that it is equal to -;;^^^^where n is the number of molecules per c.c. and d is
the d»ameter of each molecule. Show that the mean free path is proportional to
P«ssure. {A.M.I.E.)
6. Derive an expression for the viscosity of a gas on the basis of kinetic theory.
{A.M.I.E.)
4
CHAPTER IX
CHANGE OF STATE
9.1. When a substance like ice or paraffin wax is heated in a
vessel, a temperature is reached when the substance begins to pass to
the liquid state. This process of conversion of a solid into its liquid
state on heating is known as fusion or melting and the temperature at
which the substance begins to melt is known as the melting point, u a
liquid is cooled, the reverse change takes place and the liquid is said to
have solidified. The change of state from liquid to solid is called
solidification or FREEZING. The temperature at vvhich freezing
takes place is called freezing point. If the pressure is unchanged a
crystalline substance always melts or solidifies at the same temperature.
In the case of some non-crystalline substances the melting and freezing
points are not equal i.e., butter melts at 32^(7 and freezes at about
22^C.
If the liquid is further heated, it is converted into vapour at a
certain temperature. This process of conversion of a substance from
its liquid to the gaseous or vapour state is called vaporisation an
the temperature at which this change takes place is called the
point of the liquid. This is also a characterstic property of the
substance but varies greatly with pressure.
The reverse process of the change of a vapour j
is known as condensation. Certain soIk^ when ^eate known
directly into vapour or gaseous state. This ^ Camphor,
as sublimation. The examples of such substances a P
aminonivm chloride, iodine etc.
therlme “ced fn* the foM sh^ws°tt t thf gt
remains constant HU the wJwleofthe ^tme gradually rises,
liquid is continuously _ heated >ts temp rature^^g
liquid is continuously neaiea. ^oil and changes
till at a certain temper^ure the ® and more of the
into a vapour state. On further heating
182
CHANGE OF STATE
183
liquid is vaporised but the temperature remains constant till the
whole of the liquid has vaporised.
The above two examples show that at the melting point and at
the boiling point heat is continuously being supplied but the
temperature remains constant. This amount of heat which is absorbed
during a change of stale without caushig a rise 'of trmperntvre is called
the latent heat. (Latent means hidden)
There are two kinds of latent heat.
(t) Latent beat of fusion.
(ii) Latent heat of vaporisation.
Latent heat of fusion. It is the quantity of heat required to change
a unit mass of a solid at its melting jyoint into the liquid state
without ehange of temperature.
Latent heat of Vaporisation. It is the quantity of geat required to
change a unit mass of a .liquid at its boiling g^oint from the liquid to
the vapour state without change of temperature.
Latent heat of fusion of ice is also known as the Latent heat of
water. It is defined at the amount of heat required to melt 1 gram of
tee at 0 C into water at 0°C without change of temperature. Latent heat
ot fusion of ice is 80 calories per gm. or 80 k. cal./kg. and latent heat
of steam is 540 cal./gm. or 540 k. cal./kgm.
The latent heat of steam depends upon the temperature at which
equation vapour state takes place and is given by the
L=538*86+0-5994 (|00—
9 3. Laws of Fusion. The process of coversion of a solid into the
of fuL'n '' following laws known arthe llws
(t) Every substance when heated begins to melt or fuse at a
particular temperature (melting point) which depends upon thrpresLre.
{«) From the moment melting begins the temoerature of
^ubstance remains constant until the tfhole of th“'’:ubsYanc°e'
definiie’amount orhU‘to®“hange i'^rom t^Tt-H
without change of temoerature ^Thic i •‘^Ud to the liquid state
fusion of the ^ubstanT heat of
Some substancL fik^lce^cont^act*^ volume on melting,
expand on melting ^^“tract whereas others like paraffin wax
decreases in volume on melting, is lowered witf rise of "
184
ENGINEERING PHYSICS
9.4 Determination of Melting point of a Solid. The melting point
of a solid is the Um.pcrafure at which it changes into the liquid state
without change of temperature. The melting points of pure crystalline
substances like napthalene are always sharply defined. But amorphous
or non crystalline substances like wax, rubber and glass pass more
or less through a plastic stage and hence do not posses sharp melting
points. The melting points of solids like wax or napthalene which
have low melting points can be found out by the following methods.
1 . Capillary tube or Opacity method. Take the substance in a
Fig. 9.1.
small dish and melt it. Insert one end of a small capiften/ tube
into the dish. Some liquid will rise inside the tube due to capillary
action. Seal the lower end of the tube in a flame and tie it to the
stem of the thermometer near the bulb, by means of a thin
band as shown in Fig. 9.1. Suspend the
capillary tube in a beaker, about JtA filled with liquid in such '^.y
that the open end of the capillary tube is above the liquid as shown in
the figure.
fNote (i) The liquid in the beaker should be such that its hoilivg point
wax and some oil or sulphuric acia t
sulphur.^ . .
Heat the liquid in a beaker from below
it. Note the temperature at which Remove the
just becomes clear and transparari _ oil the time Again note
flame and allow the liquid to cool, stirring it
the temperature at which the melted solid ]ust oeg
CHANGE OF STATE
185
solidify i.e., it becomes opaque. The mean of these two temperatures
is taken as the correct melting point of the substance.
2. Cooling curve method. The fact that during the process of
conversion of a solid into a liquid state and vice versa, the tempe-
rature remains constant, is used
to find the melting point of a
substance. Take some substance
such as paraffin wax or napthalene
in a small vessel say calorimeter
or a beaker and heat to a tempe-
rature till the whole of it melts.
Place a thermometer and a stirrer
in it and remove the burner. .Allow
the liquid to cool and go on
recording it temperature s&y after
everj' minute. When the liquid starts
solidifying, stop stirring but go
on noting the temperature till the
substance has cooled much below
the solidifying point. The graph
between time and temperature is
plotted, the shape of the cooling
curve thus obtained is shown in Fig.
Fig. 9 2.
9.2.
The part AB of the curve shows the liquid state of the
being cooled. The temperature goes on falling
till the point B is reached where the solidification sets in. Now the
temperature remains constant till the whole of the liquid has
solidified as shown by the part BC of the cueve which runs
I'®*. t'me axis. During the part BC the substance is
liquid states exist together until the
solidification started at point B ends at point C, the t^mn^ature
rSu i.f ^hows the solid ™^ rwUh
It iV temperature as the solid is being cooled. The temD-rature
cor.espondmg to the part BC of the curvi is the melting Sof the
substance. This method is generally used for finding tn.. i+*
pomts of metals, alloys an^d crystLirne solids as I curte
definite stage of constant temperature is obtained with these. ^
in a’ *purf"‘’sttr\”" ”a“pTfLtry‘”"iea^;‘^^
turbance can be lowered to a ^ ^
fr<=®«ng point without solidifying For in^fi”" “’•1
distilled water is taken in a elas «5 * v* o instance, if
without stirring it, then it can bf coolerupto - 1 wUh t
being formed. Water has been cooled to ^ any ice
surrounding it with a liquid of the l^fd^nsit^
186
ENGINEERING PHYSICS
point. Substances like benzene, molten hypo (Sodium thiosulphate)
napthalene, phosphorous and antimony have similarly been cooled
much below their freezing points. The cooling of a liquid below its
freezing point is called super cooHag and the liquid in this condition is
said to he super cooled.
The phenomenon of super cooling does not really go against the
fact that a liquid solidifies at a definite temperature, which is its
freezing point. For, it is essentially an unstable phenomena, because
the addition of the smallest quantity of the solid or even impurity
like grit or dust particles or any mechanical disturbance such as
shaking, stirring etc. is enough to start solidification.
An interesting case of super cooling is offered by sodium thio-
sulphate known as hypo. If this is melted and a cooling curve obtained
in the ussual way, it is found that the temperature falls quite steadily
for a considerable time, obeying Newtons' law of cooling. Suddenly
solification starts, and immediately a considerable rise of
temperature takes place, temperature rising to, and remaining steady
at the true melting point until all the hypo has solidified. It then
starts to fall again according to
the ordinary law of cooling as
shown in Fig. 9.3. Thiis case
provides a very good example
of the latent heat set free
when a solid freezes. The super
cooled liquid retains this heat
until solidification occurs, but,
at the moment freezing sets
in, the latent heat is given up
and a rise of temperature may be
observed.
^ TIME.
Fig 9 .3
9.6. Effect of pressure on melting point- The P^^t of
a solid is found to depend on external pressure. The effect is as
follows.
Substances which contract on melting e.g.. ^e, kon and
antimony etc. have their melting points
pressure^ whereas those substances which expand on melting
Seir reiting points raised e.g., paraffin wax, ghee, copper etc.
The magnitude of the change in the melting point due to changes
in pressure can be calculated from the Clatisius Clapeyron equation.
dP JL
dT "TCTa-Ki)
Where d3’=change in the absolute melting temperature T
CHANGE OF STATE
187
rfP=change in the applid pressure
rj=Specific volume of the substance in the solid state
F 2 =specific volume of the substance in the liquid state
and !,=■ Latent heat of fusion.
From this relation it can be readily deduced that the melting
point of a few substances which increase in volume on solidification
or contract on melting like ice, iron, antimony is lowered by increase
of pressure, as shown below.
In the case of ice fiance dPjdT is negative. It shows
that with increase in pressure, melting temperature is decreased.
The melting point of most of the substances, whicli contract on solidifi-
cation (or expand on melting) like wax, copper etc. is raised by increase
of pressure as shown below.
In the case of wax, hence dPjdT is positive, showing that
when dP is positive dT is also positive. In other words, melting
temperature of wax, increases with increase in pressure.
Calculation shows that in the case of ice, the melting point is
lowered by about 0 0073®C per atmosphere increase of pressure,,
whereas in the case of paraffin wax, the melting point increases by
about 0'04®C' per atmosphere.
9*7. Regelation. The fact that the melting point of ice is lowered
by the increase of pressure can be shown by the following experiment.
K£ SLAB
Press two pieces of ice together for a few seconds. On removing
the pressure, the two pieces of ice stick to each other to form one
piece. It is because the pressure lowers melting point of ice, so that
the ice melts at the surface of contact. On releasing the pressure the
melting point rises again and the water
formed again solidifies resulting in the
formation of one piece. This jykeiwnienon
dccordiiifj to which ice melts whai
pressure is increased and again gets
solidified when pressure is removed is
called Kegelation (re-again : glare,
freeze). It can be convincmgly
demostrated by the following simple
experiment. Take a slab of ice and
rest It on two supports. Hang over the
ice-slab a loop of copper wire from the
that suspended as shown. It is found
‘--a.
Fig. 9.4.
is un^^ncreased ^ tL^e ^elow the wire
necessary for melting is taken from the wi^: a^nd^^the^^:
188
ENGINEERING PHYSICS
it^, and thus results in the fall of temperature of ice to a value below
O^C. The wire sinks through the water formed from melted ice and
presses on a fresh part of ice which also melts under increased
pressure. The latent heat for its melting is taken from the water
above the wire, because this water being no longer under pressure,
freezes as soon it comes above. During freezing, it gives out latent
heat which passes through the wire to the ice below and melts it. This
process continues until the wire cuts its way completely through the
block of ice, leaving the block as a single united piece as if it had never
been cut thiough.
It also follows that a metallic wire required for this purpose
should be a good conductor of heat. Thus copper wire will cut its way
throng the block of ice more quickly than an iron wire of the same
as iron is comparatively a poor conduotor of heat.
Snow balls, can be made by pressing fresh snow with hands. The
small pressure applied is sufficient to melt particles of snow inside and
removal of pressure binds the particls of snow into a stable form of
a ball. Tf, however the temperature of the snow is much below the
freezing point, such balls can not be formed, since hand pressure is
insufficient to cause melting. Figures in snow can be built on the
same principle.
Skating. The edges of the skates, being very fine, exert a very
large pressure on the ice which consequently melts. The water thus
formed acts as a lubricant and enables the skates to glide along the ice.
The edges usually sink far enough to enable them to “grip” the ice
which would not be possible if the ice is very cold and this is why
skates do not "bite" in very cold weather.
Glacier motion is also attributed to this phenomenon. The layer
•of ice at the bottom of glacier is subjected to a pressure due
weight. The ice thus melts aud water oozes out and resolidifies.
The hlocl' of ice thiLS moves continuously.
Metal casting. An important practical application of metals that
expand on solidification is made in solid casting. In order to obtain
sharp casts from a mould, iron which expands on solidification is used.
The casting in copper or silver is much less sharp because these metals
■contract on solidification.
9-8, Freezing Point Of Solutions. When »
in a pure liquid, the substance is called the P ,
called the solvent and the resulting mixture is called the solution.
Experiment shows that the freezing point of a is al^s
lower than tiuit of the pure solvent. This
■of as the depression of the freezing point. It was =hown b> Blagden that
the lowering of the freezing point of a liquid
concentration of the solute. This is known as Blagden s Law.
Later on it was found by Rauolt that for a molar
depression of the freezing point of its solution ts proper
CHANGE OF STATE
189*
concentration of the solute irrespective of the nature of the solute disoolreii .
Rauolt s Law may be expressed in mathematical symbols in the form
f\T=KC
Where A^=l^epression of the freezing point below that of the pure
solvent. *
C=Concentration of the solution in gram molecules of solute per
100 gm. of solvent. '
A^=Con.stant for given solvent and is known
constant. For substance dissolved in water the value
18-5°C.
as cryoscopic
oi f\T is about
9.9. Mechanism of freezing mixtures. .Any two sub-
stances which on mixing produce cooling constitute a freezintr
mixture. A mixture of ice and common salt (sodium chloride /KaCh
is a good example of a freezing mixture The fall in temperature is
due to the following reason. v ii.
The ice first melts and takes the latent heat of fusion from the
salt and the surrounding ice The salt then dissolves in the melted
ice and takes the heat of solution from the mixture and thus a fa I .Tf
temperature takes place. When ice and common salt are mixed In
P"°P"5or'’Th ’ •’y ‘'^■"P^rature falls to aboit
-23 6 C. This proportion of 3 : I is called the eutectic proportion
Another mixture commonly used is obtained by mi.xing calcium
^ ^ ".wi ..d
by -n * ' ■
Let us consider in further detail the freezing
solution of common salt (NaCl). When this dilutf so^IuUon of cS^on
Fig. 9.5.
salt and water is cooled down to 0 °O. it remains
liquid. When
190
ENGINEERING PHYSICS
■cooling is continued further between 3® and 4® bplofo zero some pure
ice separates from solution. This temperature at which ice first appears
IS called the freezing point of the solution. If the solution is further
cooled then more ice is formed and the remaining solution becomes
progressively richer in salt. ‘This continues until a temperature of
about — 23®Ois reached, when the whole of the remaining liquid
freezes into a solid mass. This mass contains about 23‘6 percent of
salt. A solution containing 23-6% of salt is called a eutectic mixture
and —23®C is called the eutectic temperature. This phenomenon is
well explained by the freezing curves for ice and salt as shown in
Tig. 9.5 The eutectic point at a temperature of — 23®C and a concent-
ration of 23*6% NaCl is clearly marked. To the left of C, the solid
deposited on cooling is pure ice and to the right of point C, the solid
which separated out is pure salt. At C both ice and salt are deposited
together t.e., the mixture appears to solidify as a whole.
9.10. Alloys An alloy is a mi.xture of two or more metals, and, in
general it has been found experimentally that the temperature at which
an alloy melts is below the temperature of the lowest of the melting
points of the constituents. In certain respects an alloy behaves as
though it were a solution of one constituent in the other. The graph
of the melting point of an alloy
of copper and silver against the
percentage of silver is shown in
Fig. 9.6.
Melting point of pure copper
isl 083 ®Cand is represented by
point A. Similarly melting
point of silver is 961®C as
represented by point B, If
to any one of these metals we
go on adding increasing amounts
of the other, the melting point of
the resulting mixture goes on
decreasing in a regular manner
until it reaches a minimum
value of 740®^ as represented by
i* COPPBR
Fig. 9.6
the point C. The composition of the alloy at this temperature is
40 % copper and 60% silver. The copper silver alloy of thi^s composi-
tion is known as Eutectic alloy and the temperature of 740 C is called
Eutectic temperature. Above this temperature either the copper or
the silver crystallizes out, according to the
of the alloy, but below it the eutectic alloy freezes (solidiBes) as
one mass.
Consider an alloy containing 20 % silver and 80% copper in the
nolten condition. If the temperature is decreased then as sno n y
B
CHANGE OF STATE
m
the vertical line LMN in Fig.9.6. it is copper and not silver that crys-
tallises out at point 31 and continues to do so till N is reached.
At iST the molten liquid left has the same eutectic composition
t.e.. 60% silver and 40% copper as at C. Below this temperature
eutectic alloy freezes in this composition as one mass. If molten
alloy contains 80^-o silver and 20% copper in the begining then silver
crystallises out first at the point E and continues to do so till F is
reached. Again at F the molten liquid left has the eutectic tempera-
ture as at C. Below this temperature the eutectic alloy freezes
enblock fas one mass). However if the molten alloy with eutectic
composition is considered at a temperature represented by poi it C and
is cooled gradually then it remains in the liquid state all through
and then freezes completely as C is reached without any previous
solidification of either silver or copper. The eutectic points of some
alloys having two metals in composition are given below.
Metals in the
alloy
3 . Copper and
silver.
.2. Lead and Tin.
3. Antimony and
lead.
M . Point (®(7)
1st metal
1083
327
630
M. Point (®C}
2nd metal
Eutectic point
CC)
740
186
2-16
■ ^
*8 required to convert 20 "
-30^0 into steam at lOO^C, ^ convert gms. of tee at
LaiejU Heat of steam^536 caUfgm, and latent heat of fusion -
calsfgm. specific heat of ic€=0-5 cal.jgm.f^C. J Juaton of tee
Heat is required as follows.
(a) Heat required to raise the temp, of ice from— 3o"C7 to
O^C^mSf
=20X 0-5 x[0-(_ 30)1
/M « . • . =20X0-5X30=300 cals
((6) Heat required to convert ice into water
at 0°C=,mL
r . „ . . ^ 20X80=1600 cals.
~20X 1 X(100— 0)
(A\ j *2000 C3ris«
(rfj Heat required to convert water into steam
at 100°C=mi'
=20X536
= 10720 cals.
192
ENGINEERING PHYSICS
Total Heat required
= 300 + 1 600 +2000 + 1 0,720
= 14620 cals.
= 14*62 Kilo-cals.
9.11. Boiling. ^Vhcn a liquid, such as water originally at room
temperature is heated in an open vessel, and when temperature has
risen sufficiently, bubbles ma}' be seen rising from the boUom of the
vesi'cl. The bubbles frequency contain some air together with the
vapour of the liquid. It is most improbable that a bubble should
form in the body of the liquid unless some foreign material such as a
dust ]>article, happen to be present. At relatively low temperatures
tlie bubbles cannot increase in size since the sum of the bubbles is less
than the hydrostatic pressure due to the atmosphere and the liquid.
Hence they keep clinging to the walls of the vessel and cannot rise
up to the liquid surface. As the temperature rises, a point is reached
due to the rapid increase of vapour pressure with temperature when
the total pressure inside the bubbles is equal to or just exceeds the
e.xternal pressure. The bubbles w’ill now grow in size and with further
ev’^aporation they will become two big and buoyant to remain attached
to the sides of the vessel. They then collapse away and rise to the
surface in quick succession. As the temperature is increased furthei%
more bubbles are formed which contain mostly liquid vapours instead
of air. They also rise to the top of the liquid and then escai)e into.
the air.
As the temperature rises, further bubbles of steam appear at the
bottom and begin to rise to the surface. Since the upper layers are
colder than the Icwer layers, these bubbles condense producing sharp
clicks which has a characteristic sound called the singing of the hettu.
Wifh the rise of temperature bubbles of steam cease to condense, the
singing sound stops and the vapours begin to escape rapidly into tne
air Since these bubbles do not collapse (break) on rising to the
surface, it is obvious that the pressure of the vapour inside them must
be equal to the atmospheric pressure. Ihe liquid is now s+d to D
boiling or in a state of ebullition. Thus it follows that a
hoil at a temperature at n-hirh its maximum vapour the
to th^ vressufe acting on its free surface It is for this reason that the
boiling point of a liquid is also defined as a temperature at which, Us
maximum vapour pressure becomes equal to the external pessure
Tts surhoe The Experiment shows that the maxir-nuni vapour pressu e
ofEamrat I 00 ”< 7 is 76 cm. of mercury which shows that water wUl
boil at lOO^’C if the external pressure is 76 cm. of mercury.
Thp +pmnerature of water remains constant so long as bp^mg
boils is called as its boiling point. The boiling M a liquid rem
constant if th^ pressure on its surface does not change.
CHANGE OF STATE
193
Thus boiling or ebullition is a rapid change from liquid to the
gaseous state. It takes place throughout the mass of the liquid at a
definite temperature called its boiling point.
Boiling with bumping. When air-free water is heated in a trIass
Irid previously been carefully cleaned with hydroflLric
acid and then rinsed with water, it is possible to raise the temperature
Xe TniLn T" ebullitZ taking
place. Indeed b\ a process of alternate boiling and conlinn to
expel dissolved air, it has been fonnd possible tS raUe the Xnl
ratnre of water to 105°C or 106'’C' If more ei-e is
mav boil at a still higher temperatnro This is clll^l
and this phenomenon of superheating like snpercooling is unstlblo"
enough fojbnbble’,: toTo™"‘ thef win gX” “nw rS-’’\^nr;i"'®'‘
-“S'. u
or other impurity. Sand or iron filhigs havc'tht^Xelfiect"'^®"''''^
Of formation
nrui^Ti^ “errnowTttr'^hr,'^^^^
respects like an elastic membrane, andlends "o contrn
the surface area and the corresponding potential rnake
possible. ^ "s potential energy as small as
When air bubble exists in a *u-
the bubble contract until the forces of tends to make
a force produced by the pressure within balanced by
2'. and the radius of bubble is r, the edges of k tension is
composing the surface are pulled together bv ^ f ^^"^i^pheres
counteracting force due to the excess nlJ ^rrrT. The
It may be estimated across the equatorial section of for
In equilibrium of area Trr^.
Th- T,
^ P-ss^e differen
no
194
ENGINEERING PHYSICS
Laws of Ebullition or Boiling. (») Every liquid begins to boil at a
certain fixed temperature called its boiling point.
{ii) At boiling point, the saturation or maximum vapour pressure
of the liquid is equal to the pressure acting on the surface of the
liquid.
iiii) From the moment boiling begins, the temperature
remains constant till the whole of the liquid has boiled off.
(iv) Every liquid begins to boil at a certain temperature which
differs from liquid to liquid but depends on the pressure acting on
the surface of the liquid.
(r) The boiling point is raised by the increase of pressure and
lowered by the decrease of pressure on the liquid surface.
(i't) A unit mass of each liquid requires a definite amount of heat
to change it from the liquid to the gaseous state without change of
temperature. This amount of heat is known as latent heat of
vaporisation.
irii) There is enormous increase in volume when a liquid changes
to the gaseous state. For example, when one c.c. of water is converted
into steam it occupies about 1600 c.c.
9.12. Determination Of Boiling point. Boiling point of a liquid is
that temperature at which the maximum pressure of the vapour is equal
to the atmospheric pressure. In other words it means that a liquid
in communication with the atmosphere boils at a temperature such
that the saturated vapour pressure at that temperature is equal to
the atmospheric prressure. This forms
the basis of a simple laboratory method
for finding the boiling points of various
liquids. Fig. 9.7. shows a piece of
appartus which is designed to illustrate
this fact.
\ J_shaped glass tube is closed at
the top of the short limb, the other limb
being open to the atmosphere. The
bend of the tube contains mercury. 1 he
space above the mercury in the closed
limb is completely filled bj' a shor
column of water. The air in that space
is expelled out before
water. The J tube along with a stirre^
and a sensitive thermometer is placed
in a beaker of oil or water. « J'®
this s orter limb, we shall
the temperature reaches the boi mg 3 ,
point, the pressure of the vapour atismg so that it s.tajids
irom this vvater is sufficient to depress the mercury
CHANGE OF STATE
195
at the same level in both limbs of the tube as illustrated in the
figure. This shows that the pressure of the vapour inside is equal
to the atmospheric pressure.This temperature is noted and is the boiling
point of the liquid.
The temperature is farther raised so as to heat the liquid a few
degrees beyond its boiling point and is then allowed to cool. The
temperature at which condensation takes place is also noted. The
mean of these two readings gives the boiling point of the liquid. This
experiment is not very easy to carry out because of the possibility of
traces of air being present in the closed limb.
This apparatus also affords a convenient laboratory method of
determining the boiling points of other liquids such as ether and alcohol.
Boiling point of solutions- Similar to the case of freezing point
■of solutions, the boiling point also is affected by the process of
solution. Thz hoilinj point of a .•iolution is hijhzr than that of the
pare solvent.
Solutions obey the same law of boiling as pure liquids viz^, the
satumted vapour pressure of the liquid at its boiling point is equal
to the atmospheric pressure ; but there is an important difference,
uiz., the vapour pressure of a solution at a given temperature is
always less than that of a pure solvent at the same temperatare.
inus the vapour pressure of a salt solution at IIO^C is less than
76 the salt solution has to be heated to a temperature higher
than 100 a for its vapour pressure to reach the value of the atmospheric
pressure when boiling will set in. ^
The elevation of the boiling point of solution is found to be
proportional to the concentratian of the solute. Raolt’s Law which
ei:vl°tio"a tl>e freezing point also holds "-good^fo?
•electrolytes and solutions of o'’rgrn‘c coLpounds.'
The law states that
*
4 A V
■concen^ation omnolu^bn^’in^gm mol ^ boiling point. C the
of solvent and K a constant for tL
aqueous solutions. soh cat whose value ig 5-2®f7 for
discussed aboutShe v?poa7pressure^^^ . From what has been
it follows that if the externll^nrf.f ^ ^ its boiling point
liquid will boil at a tvh- ^ is alterfd^°the
will be equal to the ifew ^ vapour pressure
an increase of pressure raises th^ u ^rV^rnal pressure. Hence
decrease lowers it. The magnitude oT ‘tte®
■due to an increase in pressure ^oiUng point
given by the Clausius-Clapfyron
196
ENGINEERING PHYSICS
thermodynamical relation.
dP_ JL
dT T{\\-V^)
where dP— change in pressure
dT=rise in boiling temperature in
Pj=specific volume of the liquid
|’2=specific volume of the vapour
7 '=boiling temperature in
J=Mechanical equivalent of heat
P= Latent heat of vaporisation.
If will be seen that when a liquid^, passes from the liquid state
to the vapour state at the boiling point, r2>^ :■ Thus ^ positive
quantity which means if dP isipositive.^dT is also positive.
9 14 The triple point. The triple point may be defirted as the
ujJ the mat ur i.e..
eyiMbrium. Ulu s at t he junctmMMin^hr,e . ue Ime and Marjr
line. ^ ,
I magine a cylinder fitted_jw ith a g^tigM piston ^atrof
a substance partly m The liquid __^at c^ and Law
i.
I
§:
U 0 U/D(^^^£^)
V.^POUR
(srmi)
O temper AT (JRE
Fig. 9.8
cantact. The relation j^'"^,s|linriiiroOina*ti1.nd teinperfture
':nr.“ata£; ^
CHANGE OF STATE
197
When the temperature of the system is changed, the saturation
vapnii r pressure is also chan ged. If a graph is plotted between
maximum vapour pressu re and temperatur e a c urve of the type P.Sf
as shown’in Fig. 9.8 is obtained. This curve is called the curve
of vaporisation or steam line in case of water and steam. All points
on the curve PS correspond to an equilibrium state between liquid
and vapour.
For any point above the line the ^bstance is entirely in the liquid
state and for any point below the line the substance is entirely in the
vapour state. A liquid and a solid similarly exist together in
equilibrium at the temperature at which the solid melts under a
given pressure. This pressure is also a function of temperature only,
and relation between pressure and temperature is represented by
the curve PI as shown in Fig. 9.9. It is called the curve of fusion or ice
line in case of ice and water. We have seen in connection with the
change of state from solid to liquid that the substances may be divided
into two classes as given below.
Fig. 9.9.
which expand on solidification e.g, ice. In such a case
the melting point is lowered by increase of pressure. This case is
represented by the curve PI which slopes towards the left, (til Those
which contract on soUdification wcoc. In such a case the melting
fi? i raised by increase of pressure. This case is represented bv
the dotted curve PJ which slopes towards the right. ^
equilibrium may exist between a
between pressure^'^and f representing the relation
Bpt S i«ts iLwn^Si,!
ENGINEERING PHYSICS
When all these three cun-es are drawn on the same diagram
Fig. 9.10.
they meet in a single point P, ^ called the triple point or the funda-
mental point as shown in the Fig. 9.1 1.
pig. 9.11.
Tt mav now he defined as the pomt where, Jor a particular
Jt may Unuid and vapour exist simultaneously %n the same
iatr'-is-r/s. ... 'bSiSr.?‘ .t s
states, but m substance is entirely in the solid state
States on the P-T f ^^ely in the liquid state in the region /PS and
...t. I" ... '•*»" ff-
Th. „ipi. p.i.. .■ . p'-~ “
fication is above its | state above the triple point. The
CHANGE OF STATE
is below its melting point under normal pressure and the substance
cannot exist in the liquid state above the triple point.
Numerical values for water at the triple point. At the triple point
the pres sure of the saturated va pour is e qual to the pressure_of fusion
and it is also equal to the pressure o f subT imafibh. The teiiTp^rature
is sligHn y^above 0^^ At 0°C the maximum vapour pre ssure of water
is 4.6 m.m.
The specific volumes of water in the three states at the triple point
are 206000 c.c. (vapour). 1 c.c. (water) ; I 09 c.c. (ice). Regnault
thought that the hoar-frost line was a simple continuation of the
steam line, but it was shown by Kirchoff that they are two distinct
curves meeting at an angle. The direction of each curve at the
triple point may be found by calculating the value of ^ by apply-
ing thermodynamic principles.
9.15. Vapour pressure over curved surfaces- Due to surface
tension the pressure of a saturated vapour in equilibrium with its
liquid is not the same when the liquid surface is curved and when it is
plane.
An expression for the vapour pressure over
surface can be obtained as follow. A
capillary tube is placed vertically in a liquid and
the whole is enclosed in a vessel from which air has
been exausted so that the space above the liquid
U filled with its vapour and with nothing else.
In general, the liquid in the capillary tube stands
at a higher level A than the level B and its surface is
curved (concave upwards) as shown in Fig. 9.12.
There is a state of equilibrium between the
liquid and its vapour both at A and By otherwise
evaporation or condensation will occur until
equilibrium is reached. We will assume that the
temperature is constant.
Let po=vapour pressure in contact with the
plane surface that is, the pressure at B.
p-Vapour pressure just outside the liquid at
A i.e. A'
a curved liquid
Fig. 9.12
Pi=*the pressure just inside the liquid at A i.e.A’
p=the density of the liquid,
a'=the density of the vapour (assumed constant)
The pressure at B exceeds that at .,4 due to the column of the
vapour of height AB. Let this height be h, then
is PJ^essure exerted by a column h of the vapours of density o-
..(t)
• •
hag
200
ENGINEERING PHYSICS
The pressure due to a column h of the liquid of density p is hpg
p^—Pi=hpg ...(u)
The curvature of the liquid surface in a capillary tube is due
to the surface tension T. The difference in pressure across this
concave surface i.e., the pressure difference between the inside and the
IT
outside of the surface at A is equal to — where
r is the radius of
curvature of the surface at A.
IT
P-Pi=- —
Subtracting (i) from (ti) we have,
p—p^^gh{p—a)
'>T
— — 0-)
or
gh{p—&)^
IT
[Hi)
If we substitute this value of gh in equation (t) we have
IT \
,-p=<T * 7 ” )
Pa
P — (T
or
P=Po-‘
IT
P — (T
...(iv)
This shows that the vapour pressure in contact with a concave
surface is lower than that in contact with a plane surface by an
^ 2T
amount .
P—cr
-depending on the radius of curvature r of the
surface.
Similarly if the surface is convex towards
above expression becomes.
2T
P-Po^—T'
the liquid thvn the
<r
Hence
P=Po+
P^(T
2 T <r
Thus the vapour pressure in equilibrium with a convex liquid surface
2T . <T
is greater than for that a plane surface by an amount-
Representing ,the difference in vapour pressure over plane and
CHANGE OF STATE
201
curved surface by we have
A 2r
Ap=~
<T
p~a
From the above expression it is clear that the pressure difference
Ap varies inversely as the radius r of the curved surface. This fact
has important bearing on the phenomenon of boiling and formation of
clouds.
(0 Boiling. In the process of boiling, if tliere are nuclei
radii due to the presence of dust particles or iron
filings or of porcelain bits placed in the liquid, the bubbles formed
round these nuclei are fairly big, so that the excess pressure due to
surface tension is of moderate amount and tlie vapour pressure can
easily support it along with the hydrostatic pressure, thereby leading
"to the normal and regular boiling.
If on the other hand, there are no or very small nuclei present,
the excess of pressure due to surface tension will be extremely large.
fo>-niation of bubbles cannot
between wilUead to superheating where the equilibrium
Ubil w?th th pressure and hydrostatic pressure will be very uns-
table with the consequent boiling by bumping. ^
{»} Formation of Clouds. Dust particles and charged ions olav
n iinportant part in inducing the condensation of water vapour in
e^la n^^n thVfoh'^" • clouds. fop„d mist etc. This'^an be
* following manner. Clouas,fogs and mists are aggregates
Th^se a 1 1 Zell?:/ general are found round dustIfrtfcTes
^ nuclei of appreciable radii of curvature so that sun^r
water no^'te b^so^r h*’" atmosphere loaded
Jd|Pr'“pZ llTjlTio a temUat^r
in contact with excess ^<^POur
60^0 are respectively^ arid """ 27^0 and
pressare of water at 27^0 is 2-7 ^ mercury. If the vapour
pressure at 60°C ? ^ mercury, what is the vapour
The pressure of the air at 27®C==77-7_2.7__7, n
The pressure of the air at 60 ®C= 9 ij-i_p
•where is saturated vapour oressurp or *
the gas at constant volume is nrnn" f • * the
temperature, we have for the constant^n;Ls'’XiJtThV*rxtum^^^
^^'^=^273 + 60
75-00 ^- 273+27
202
engineering PHYS»CS
or
98.1_^^lii-X75-00.
300
jj=14‘85 cm.
Example 3 What n^s of "o.er
freezp 5 grams of water at 0 C i n j r
is 95 caJjgm. and that of water is 80 calsfgm.
Mass of water to be frozen. m=5 gms.
Temperature of water =0®C.
Latent heat of water L=80 cals/gm-
Heat lost by water to form ice=fflX-=5 X 80 . =400 '
Let the mass of ether evaporated to produce enougi c
freeze 5 gms of water at O^C^ be 3/ gm».
Latent heat of vaporization of ether=95 cals/gm.
. ar V QS cals,
Heat gained by ether
But, Heat gained
or
=3rx95 cals.
=Heat lost.
37X95=400
= gms.
95
/ T^ent O f) contains
Example 4. -4 cahrim-Jer
VZlre until >ke ur.pe^^ Xafent
steam condensed ^Latent heat of Jusion oj
“ r "-f .. «.
of ice)gets heated up to 10 C. f 50 y 80 )+( 50 -l- 50) X 10
Heat required for this ca\l
r it rr ^
’"ns .. ...» “ '
cools down to >0 en f ( 100 - 10 ) cals.
630 37=-5040
Now ,i^_<nao/630=8 gms.
CHANGE OF STATE
203
9 . 16 . Vaporisation. When a liquid is heated it changes into vapour
form. The change from the liquid to a vapour state is called
vaporisation- The reverse phenomenon t.e., change of vapour into
the liquid state is called liquefaction or condensation. Vaporisation
may take place in two ways.
(t) Evaporation. In this process there is a slaw formation of
vapurs at all temperatures and takes place only at the surface of the liquid.
The gaseous state of the liquid is called the vapour. The liquids which
readily evaporate are said to be volatile and those which do not
evaporate at ordinary temperature are called non-volatile.
(it) Ebullition or Boiling. In this process, there is a rapid
formation of vapours from all parts of the liquid and at a constant
temperature.
Factors favouring evaporation. The rate of evaporation is found
to depend on a number of factors.
(i) The nature of the liquid. The lower the boiling point of a
liquid, the quicker is the rate of evaporation. If the liquid, is more
volatile the evaporation will be rapid. For e.Kample.alcohol evaporates
more rapidly as compared to water and ether still more rapidly than
alcohol.
(n) The area of the exposed surface. As a general rule the larger
free surface of the exposed liquid more rapid is the evaporation.
Thus if we take equal quantities of water in a wide dish and in a
bottle, we find that water in the dish evaporates much earlier. It is
for this reason that volatile liquids are kept in a narrow necked bottle
which is corked tightly.
fm) The temperature of the liquid and air. If the liquid is at a
higher temperature, it will obviously get vaporised quickly.
{tv) The renewel of air In contact with the liquid surface Tlie
w-t, ^ become almost
K ^ ' vapour in a short time and hence these cannot
'® renewed, the evaporation will
With the liquid is removed quickly
! 1 very quick. This is why evaporation is moJe
blowing. ^ ^ «'ind is
C*’) The pressure on the snrface of the liauid The iecc«T. tk
pressure on the liquid surface, the lower is the bdlintr
hence quicker is the evaporation. ThuHn vacuum
IS extremely rapid. vacuum, the epavoration
{vi) The dryness of air. The drier the air th^
amount of moisture required for its saturation’ greater is the
204
ENGINEERING PHYSICS
be liquified only by the increase of pressure without lowering its
temperature, e.gf.. water vapour, alcohol vapour etc. On the other hand
a gas is that, to liquify which pressure on it has to be increased and
also its temperature has to be lowered e.g. oxygen, hydrogen etc.
For every substance in the gaseous state there is certain
temperature below which it can be liquified by suitably increasing the
preseure, but above which it cannot be liquified however great the
pressure may be. This temperture is known as critical temperature tor
that substance. Hence in true sense a vapour is a gas above %ts cr%mat
temperature and gas is a vapour below its critical temperature, tor
example, carbon dioxide should be called as a vapour at its room
temperature (25®C) because its critical temperature is 31 C.
9 18- Vapour pressure. As already described, a liquid gives off
vapour from its free surface at all temperature. As a vapour '5 also the
gaseous state of liquid, it is expected to exert pressure m the same
way as a gas does. A vapour actually exerts pressure can be
proved by the following experiment.
Experiment- Take two barometer tubes -4 and B
xnetre long and from 0-5 to 1 cm. in diameter. Fill them wi p
The height of
indicates the
is called the
Fig. 9.13
dry mercury and close their open ends mercery
Atmospheric pressure. The space above mercury
introduce a drop of eth o the surface
:i -p'~“ “
CHANGE OF STATE
205-
the tube as shown. This shows that the vapour formed
exerts pressure. This pressure is meassured by the depression of
the mercuey column caused by the vapour. Introduce a little more
quantity of ether into the tube B. It is found that ether again
vaporises, but not so quickly as at first, and the column of
mercury is further depressed. If the ether is continousiy introduced
a stage reaches when it does not vaporise but appears on the surface of the
mercury in the form of thin layer of the liquid. If now more ether is
introduced, the mercury column will not be depressed any more. At
this stage, the space above the mercury is said to be fully saturated
and contains maximum amount of vapour and is not capable of holding
any more of the vapour at the given temperature. Thus, when a space
contains the maximum amount of vapour under a given temperature, the
vapour is said to be saturated and the pressure it exerts is called the
or saturation vapour pressure h. It is usually referred to
as S.V.P.
If the vapour present in the space is less than the 'triaximum that the
space can hold at that temperature, the vapour is said to be unsaturated
vapour. The pressure exerted at this stage is called the unsaturated
vapour pressure-
9.19. Properties of saturated vapour. The saturated vapour e.xerts
maximum pressure at a given temperature in a closed space and
generally it exists in contact with its own liquid.
(i) The maximum vapour pressure is different for different liquids
(n) The maximum pressure for the vapour of a liquid at a given
temperature is independent of the volume of the space.
(lit) The maximum vapour pressure depends upon the temperature
as shown in the Fig. 9.14. From the figure it is clear that
(а) The maximum vapour pressure decreases as the temperature
falls. ^
(б) The maximum vapour pressure increases as the temperature
PI s^s*
Fige 9,14.
Note* “1 he curve is conc&ve udw^tHc *
that maximum vapour pressure increases rapfdly
'■206
ENGINEERING PHYSICS
9.20. Behaviour of saturated and unsaturated vapours with the change
• of (a) Volume (b) Temperature.
Consider the vapour of a liquid enclosed in a space above the
• mercury column in a tube B as shown in Fig. 9.15. To cause a change
in the volume of the vapour, the mercury tube can be moved up and
down while the change of temperature can be produced by heating or
cooling the vapour.
1 . Ejfect of the change of volume on vapour pressure at constant
temperature.
Again set two barometer tubes, A and B in a mercury trough as
shown in Fig. 9.15. Introduce into B a few drops of Water or ether so
as to have unsaturated vapour in the space above
mercury. The volume of vapour and also the
difference between the levels b and c of the
mercury column in the two tubes which gives the
pressure exerted by the vapour are noted. Now
lower B a little into the mercury trough as shown,
the volume of the space occupied by the
vapour will decrease. As the volume is decreased
the mercury column in the tube B is depressed a
little more, which shows that the pressure exerted
by the vapour has increased. Go on lowering the
tube gradually and number of readings for pressures
and volume of vapour are taken and is found that
at every stage, the product of pressure and volume
remains constant which shows that an unstaurated
vapour behaves like a gas and obeys Boyle s law.
Fig. 9.15
If the volume of the vapour is
the tube down into the then a S temperature,
above mercury becomes saturated with P. , . ^ ^ of mercury
At this stage, the difference h that tempe-
in the two tubes gives the maximu^m p P mercury trough
rature. Any further lowering of the tube B into me^
does not depress the mercury column a ^ condensation.
Bo,.e's Law so
Inna as it IS in its unsaturated stage.
TK, b.h..io„ .r . v.p..r wh.. 1.. £ “5
stage. At the point B corresponding to
saturated and exerts ^^condense and the pressure
. decrease in volume cause some vapour to
CHANGE OF STATE
207
remains constant. This is shown by horizontal line BC. Thus satu-
rated vapour does not obey Boyle’s Law.
Fig. 9,16.
^ JPff f Fig. 9.17.
2. of temperature on vapour pressure at constant volume.
-r- ■ -
thetubeBsoastoenclo4 some *^to
above mercury. The volume and
the two tubes is noted and is the mea^ure^/Jhe'tapo^r^^lsi^:!^
^e in the outer jackeTand keep the'^tSlumeXf ‘^fh"®
A decrease in temperature will increase k constant.
column a little which shows that the nressnre » j mercury
has decreased. If the temperatme is ^he vapom
measured at different stages theTATs see^?K pressure is
kept constant the pressL is alwlts ®re ‘he volume is
temperature. ^Ais sAouw P^Pftt.onal to its absolute
when the vapour is able to ff(z/Mra/P ^ is reached
exerts maximum pressure at that tenfn?^!*^ space fully and therefore
causes the vapour^ocondense‘^d'47;i"^^^ farther 0 ^ 01^1
vapour pressure wiU begin to
20S
ENCFNEERriVG PHYSICS
I
P
A
B
fall more rapidly with fall in temperature. This shows that the satu-
rated vapour does not obey Charle's Law.
The above change in the vapour pressure, when its temperature is
varied at constant volume is repres-
ented by the curve shown in
Fig. 9. 18 . During the part AB of the
curve the vapour is in its unsatu-
rated stage and any decrease in
temperature produces a proportionate
decrease in pressure, so that Charle's
Law is being obeyed. When the
temperature decreases to B, the
vapour becomes saturated. Any
further decrease in temperature
causes some vapours to condense and
the pressure begins to fall more
rapidly as shown by the curve BC.
Which shows that ft does not obey
Charles Law. Fig- 9-i8.
9.21. Dalton Law of partial pressures. So far we have been
dealing with the case of a particular vapour in a closed space. Now we
shall study the behaviours of a mixture of vapours and can be stated
as follows.
(i) The saturated vapour pressure in a closed space at a given
temperature, depends only on the temperature and is independent of the
pressure of other vapours or gases having no chemical affinity for each other.
(ii) The total pressure exerted in a closed space by a mixrture of
vapours and gases having no chemical affinity is the sum of the partial
pressure which each could exert if it alone occupied the whole space ai the
same temperature.
The first law is applicable to saturated vapours and the second
law to both saturated and unsaturated vapours.
9 22 . Kinetic Theory of Saturation. If at a given
r :tTe“hr Hq^uid -a the vapour
unt'il no more liquid remains, and unsaturated
nm^more liquid is if and The 'va^pouHs
5‘sr..p':r& sj’-t.irJ's-r.....
CHANGE OF STATE
209
temperature, no matter how much liquid is introduced into the closed
space. This is called the "saturation vapour pressure" of the liquid at
the given temperature. At this stage the density of the vapour is
increased to such a value that the number of molecules leaving
the liquid surface per second is the same as the number of molecules
returning per-second to the liquid from the vapour. Thus a state of
dynamic equilibrium exists beUveen the molecules which constitute
the vapour and those which constitute the liquid. Further introduc-
tion of the liquid into the closed space does not make more molecules
to leave the surface per second. It is due to the fact that the speed
of the molecules depends only upon the temperature of the liquid.
Thus the dynamic equilibrium is maintained with the same average
number of molecules of vapour above the surface of the liquid.
Therefore this vapour reaches the maximum possible density at the
given temperature and is It exerts the maximum possible
pressure, known as saturation vapour pressure. On decreasing the
volume of the space above the liquid more molecules of vapour
return to the liquid per second than in the previous case. But the
number of molecules leaving the liquid per second remains the same
and as a result some quantity of vapour condenses until dynamic
equilibrium is again restored. In case the volume of the space above
the liquid is increased, the density of the vapour decreases and thus
it becomes unsaturated. The density of tlie vapour again reaches
Its maximum value by the evaporation of some more liquid and
dynamic equilibrium is once again attained. Thus at a given
temperature of the liquid, the saturation vapour pressure remains
According to kinetic theory of matter, the molecules of a liquid
fit the liquid the mean
velocity of the molecules increases, which, therefore runabout more
rapidly within the liquid. At certain temperature some of the mole-
kinetic energy that they leave the surface of tL
iquid gainst the attpctive pull of the liquid molecules. It is these
escaped molecules which constitute the vapour of the liouid ab^ve
the surface. Further increase in temperature^ of the
the number of molecules leaving the surface Der.<iPrnn3
constantly collide with^each
other and undergo changes m their velocities and directions
molecules exert pressure when they strike the wtlU of f ho ^
the liquid surface. The vapour molecules wMcVsU ke the
do not get reflected back blit aro surface
liquid tie numb^rl" ™ll"e“l,erLalir/‘’aud Pe tuS%o
surface increases but is not the same Tn « ^ to the liquid
soon reached when this space cannot accommnHit ^
molecules. At this temperature the nuXr ol vapour
returning to the liquid becomes \
“nr
210
ENGINEERING PHYSICS
vapour is maximum. This maximurr pressure is called saturation
vapour pressure.
9-23. Moisture in the air. Evaporation of water continuously takes
place at all temperatures from rivers, lakes, oceans and from other
wet bodies exposed to air. Animals and plants also constantly give
moisture to the atmosphere. Millions of tons of water are thus drawn
into the atmosphere every hour. As a result of this the atmosphere
always contains some water vapour. Sometimes it contains only
a little moisture and has the capacity of absorbing much more of it.
At other times, it contains a considerable amount of moisture and
can absorb very little or no more. A knowledge of the amount of
water vapour or moisture present in air is necessary for the purpose
of fore-casting v/eather or for air conditioing.
The branch of Physics which deals with studying and measuring
the amount of water vapours (dampness) in atmosphere is called
Hygrometry.
Humidity of air. Presence of water vapour in the air is the cause
of humidity of air. It may be measured in absolute or relative terms
and in the two cases it is respectively known as
{<) Absolute humidity (ei) Relative humidity.
vO Absolute humidity. Dryness or dampness of air is not only
determined by the amount of water vapour actually present in
certain volume of air but it also depends upon the amout of
water vapours required to saturate the same volume of air at the
same temperature.
Let 20 gm. of water vapours be contained in one cubic metre of
air at 25°C, when 22-5 gm. of water vapours are required to saturate
air at that temperature and let 17 gms. of water vapours present in
the same volume of air at 20®C' when 17.1 gms of water vapours are
required to saturate the same volume of air at the same temperature.
In the second case though amount of water vapours present in the
same volume of air is less yet water vapours are nearl3" saturated
because less amount of water vapours are required to saturate air a
Absolute humidity is defined as the ^ amount of water*
vapours actually present in one cubic metre of air. Its units are gm/
cubic metre.
(ii) Relative humidity.To get a correct idea of the degree of saturation
of air, we must know not only the amount of water vapour actually
present in air, but also the amount of vapours required to
saturate the same volume at the same temperature. The ratio of these
quantities is called the relative humidity or simply the humidity.
Me\iiti\ehumiA\iy is defined as the ratio of the tmss of water vapour
actually present in a certain volume of air, to the mass of water vopouf
required to saturate the same volume of air at the same temperature.
CHANGE OF STATE
211
If m is the mass of water vapour actually present in a certain
volume of air and M the mass required to saturate the same volume
at the same temperature, then
Relative humidity, R.H.=
m
M
As water vapour obeys Boyle's law up to the saturation stage
the density of vapour and hence its mass is proportional to the
Relative humidity —
pr'eTsufe ^ -P°-
percJnt^ge"'’'*"" - '^action or as a
temp^^afurrftnstftTci^ar tha?t“hJ - ‘he
as the temperature falls, or decreases Ts th'e™t'*‘*^ increase
^vithout any change in the density o7water vapour p?e™se"t
not enougrto‘’sIturatr‘"and the‘‘'n generally
■water vapour is less than fh ® Pleasure actually exerted by the
temperatuT men t^e vapour pressure for^tJat
temperature is reached at which^thr actua? ^
vapour contained in that portion of water
A slight further cooling will cail^e ^ it.
5s not capable of holdfn^ to ^ the air
temperature at which thp condensed into water. This
tleiv point and it may be defined ^."" 0 ^" ‘he
Dew point is that temperature at whirh tho »
vapour in a crtaiu volume ^
dew pVint 'rs\V«rtl"\h?°pTesfu:: ‘o^* tr^^1 P— « at
present at the room temperature. Hence the vapour actually
•be debned as follows, relative humidity may also
S^^u^tionj ^ pressure at de w point
tinsa?qied'^il‘/?ooTs"d^!^^"X^^^^ ^ '-f volume of air in the
whole of it becomes saturated Th» * ^ stage is reached in which
.man dust particles and wt‘|e^i so^n^^^?oS=
, ENGINEERING PHYSICS
mist is ordinarily called fog. Fog is generrlly formed when a mass
of warm air becomes sufficiently cooled by coming in contact with the
cooled surface of the earth during the evening, night or morning. In
autumn or winter, fog is often seen on the water surfaces also.
The mist or fog formed high up in the air is called cloud.
Moist air is lighter than dry air. When air after absorbing
moisture from the sea surface rises up in the atmosphere it
becomes cooled due to following reasons.
(i) the moist air on rising comes in contact with cooler air in
the higher regions of atmosphere.
(it) The expansion of air as it moves up into regions of lower
pressure causes cooling.
When the temperature of the ascending air as a whole falls below
the dew point the excess of water vapour immediately condenes on
floating dust particle as well as on the salt particles coming from ttie
sea along with the air, thus forming a cloud.
Examples. On a certain day the dew point is and tempe-
rature of air is Find the relaiive humidity.
vapour pressure at 8°C, 9’-C. 18°C and irC are respectively 8-04, 8 6K
16-46 and I6'46 m.m. of mercury.
Dew point =8*5 o
max. vapour pressure at S°C—B‘04 mvi
max. vapour pressure at 9®C=8-61 mm
Increase in vapour pressure for TO when temperature increases-
from 8° to 9°C'=8-6I-8-04 = 0'57 mm
Increase in vapour pressure for 0*5®C7
= 0*57 X0*5=0*285
Max. vapour pressure at 8*5‘’C=8'4H-0*285 — 8*325 mm.
Temperature oTair =I8*4®C
Max. vapour pressure at 18®C-=15*46 mm.
Max. vapour pressure at 19°C=16'46 mm.
Increase in vapour pressure for 1°C when temperature increases
from Ig^^'to 19 “C= 16-46— 15*46=1 myu
Increase in vapour pressure for 0*4®C'=I X 0-4=0 4 mm.
Max. vapour pressure at 18-4°C'=I5-46 + 0-4=15-86 mm.
Relative Humidity
Afaximum vapour pressure at dew^pomj: — ^ 100%.
^ Maximum vap our pressure of air
_ Max imum vapour p re ssure at y ioqq^
Maximum vapour^ressure at 18-4®(7
CHANGE OF STATE
213
Example 6. 40 litres of air at 18^C are passed through drying tubes
and the increase in weight produced ia 0‘S96 gm. Having given that 15'2
gms. of water vapour is required to saturate 1 cubic metre of air at
determine the relative humidity of air.
Increase in weight =0‘396 gm.
Volume of air passed through drying tubes
»40 litres
Amount of water vapour actually present in one litre of air,
0*396
—0*0099 gm.
Amount of water vapour required to saturate 1 cubic metre of air
at the same temperature=I5*2 gms.
Now
1 Cubic metres
lOOxiOOx 100
1000
000 litres
Mass of the water vapour required to saturate l litre of air at the
same temperature,
15*2
1000
= •0152 gm.
Relative Humidity
X 100 =
0*0099
0*0152
XI 00
= 65*1%
Example 7. What is the degree of saturation in air at 16’^C when
^o^^ationof moisture takes place at 10^0 ? Pressure of water vapour
at JO o—ij-o mm and pressure of water vapour at 10^0^9-2 mm
Since the condensation always takes place at the dew point
Dewpoint ' s=l0®C
Max- vapour pressure at dew point p =9 2 mm
Room temperature
Maximum vapour pressure at room temperature P=l 3-6 mm
Relative Humidity 9*2
— ■p >^100 =jj:^xioo
ENGINEERING PHYSICS
il4
Saturation vapour pressure of water at 5^C=6'6 mm and at
20°C=J7‘5 mm.
i emperature at which dew forms
Temperature at which dew di3appeares=5‘4°C7
Dew point ^ 4 6+S 4 _^o0
Saturation vapour at 5^C
=6*5 m.m.
Saturation vapour pressure at 20®C =17.5 m.m.
Sat. V.P. at dew point
“ Sat. V.P at room temp.
65
Relative humidity
X lOO
17-5
X 100 = 37.1%
Example 9. The temperature of air in a closed space is obtained to be
and the dew point is 8°C. If the temperatare falls to 10°O h/>w
(i) ike pressure of aqueous vapaar in air and (ii) dew point are affected.
Oive7i pressure of aqueous vapour at 7°C=7-48 m.m. and at 8°C=8‘2 m.m.
The temperature of air in the closed space is 15®C' and the
dew point is 8®C, hence the vopaur in the closed space is unsaturated
which obeys Charles’ law.
Now pressure at 15''0=M:ix. Vapaur pressuse at dew point 8°C.
^8 02 m.m.
Let P be the pressure of
Chaale’s Law
vapour at 10°C, then according to
p 8-02
27H^^ 273+15
8-02X283
288
= 7*88 m.m-
Example 10. Calculate what fraction of the mass of the
condense if the temperature of the air falls from 20^C to 5 G
ZiTif orilinally the humidity uus 60% {Saturaiion vapour pressure at
2 Q'^q—J 7,6 m.m. and at 6°C — 6'5 m.m.)
Relative humidity =60%
Let p be the actual vapour pressure at 20°C
Saturation vapour pressure at 20“O=17-6 m.m
We know x lOO
or
CHANGE OF STATE
215
^>=I7*6X y^=IO*56 m.Trt.
Maximum vapour pressure at 5^C=6'5 m.m,
The temperature falls from 20^C to 5°C, therefore, the vapour
pressure falls from 10 56 m.m, to 6-5 m.m.
Fall in pressure due to cendensation
=ai0‘56 — 6*5=4*06 m.m.
Fraction condensed=— = 0*384
10 56
air Mo/ ^ the relative humidity of
O and J C are 17 0, 9 2 8-6 m.m. of mercury reepicUvely .
Relative humidity at 20°C=52% =0*52
Saturated vapour at 20®C', P=i 7*5 mjn.
Let P be the actual vapour at 20®C.
Now R.H. ^ ^
.*• xF=0*52x I7*S=9-1 m.m
9*1 temperature at which actual vapour pressure
9 1 m.m. becomes the saturated vapour pressure. pressure
Now saturation vapour pressure at 9®C=8*6 m.m.
Saturation vapour pressure at 10 ®C=a 9.2 m.m.
of I«c i" of 0-6 corresponds to difference
Required difference of pressure=9-2-9-f=0-l m.m.
Hence difference of O f m-m. will correspond to a difference of
temperature of ^-^^X 0 .I=-L =o i66*C
Dew point = io*0— 0*I66==-9*834°C
Onmhat ^sumptioZ ZZu;
presmre of to£er vapour at ^ Saturation vapour
^17.86 m.m. ^ JO o =72-67 m.m. of mercury, at h^C
Now 72 fT ^ Actual V.p. at room temp
S<xt. l-'.iJ. at room temp.
or Q.g P
12*67
i>-0*6XI2*67«=7*60m.m.
we can'Stes‘the‘’4l“Tact?a1 vXrVretu";:
216
ENGINEERING PHYSICS
Now
or
^)i=7‘60 m.m.
rx==273 + 15=288°^
T2 = 273+2a.-=293°A'
P2=1
Vx -_P2
A
3>2 =
or
760
" 293
7 -60X293
288
288
= 7-734 m.m.
Relative humidity at 20®^
7*734
Example 13. Find the mass of a litre of moist air at a temperature
of 32^C and pressuri of 758*2 m.m., the dew point being 15°t\ The
saturation pressure of aqueous vapour at 32°C is 12'7 m.m. The density
of dry air at N.T.P. is 1.2u3 gmjlitre'
Pressure due to water vapours=12*7 m.m.
.*. pressure due to dry air aIone=a758*2 — 12-7 =745*5 m.m.
Mass of 1 litre of moist air at 30°(7 and at a pressur of
758-6 m.m.=5mass of 1 litre of dry air at 32®C and at a pressure of
745*5 m.m.+mass of l litre of water vapours at 32®C at 12*7 m.m.
pressure.
Now volume of 1 litre of dry air at 'il°C and 745*5 m.m.
273 745*5
when reduced to N. T. P. = i ^
=0-877 litre.
Volume of 1 litre of aqueous vapour at 32®(7 and 12*7 m.m.
27 3 12*7
when reduced to N.T.P. = 1 X ^ X-i^ =0*0149 litre.
mass of dry air =0*877 x 1 *293 = 1 -136 gm- and mass of
aqueous vapour contained =0*0149x 1-293x0-622=0*1203 gni.
Note. [Assuming that]the density of aqueous vapour with respect
to air is 0*62 2.]
mass of 1 littre of moist air=ri36 + 0-01203 = l*14803 gm-
9.25. Hygrometers. Literally hygrometer means an instrument
for measuring moisture, or humidity because hygro nieans n'o^st’^e.
There are a variety of hygrometers which may be broadly classinen
as,
(i) Absorption hygrometer such as chemical hygrometer.
[ii) Dew point hygrometer such as Daniell's and Regnault s
hygrometer.
CHANGE OF STATE
217
(in) Empirical hygrometer such as wet and dry bulb hygrometer
and hair hygrometer.
(i) Chemical hygrometer. It is the most direct method of deter-
mining the absolute humidity of air. The apparatus consists of two
U tubes A and B containg some hygroscopic substances like
phosphorous pentaoxide or calcium chloride crystals. The tubes are
PumtcE STOues
Fig. 9.19
connoted by a small rubber tubing E to an aspirator C through a
consists of a bottle containg some pumice stones
soaked in sulphuric acid and serves to prevent any moisture from
aspirator getting back into the tubes A and B. The open end of the
tube A IS closed with a plug of cotton. ^
the tubes ^ and Bare
^eighed along with their contents before they are put in position
JoUon aspirator is opened and simultaneously the
the air^Sfom '^ater rushes out of the aspirator
from the atmosphere is sucked into the IT-tubes its
moisture is absorbed by the hygroscopic substance The water
coming out of the aspirator is collected^in a measuring cyUndrr to
plug is replaced^ thp- TJ ^ ^ cotton
diffience'^in the initiafl^d final again. The
vapours present in the vnW f ® j amount of water
H£MIDlTY"‘.Thl":m"ctT:ftaTer‘'vr;o^^^ ABSOLUTE
or io«c.c. of air is calculated. apour present in a cubic metre
required to satumtf ?hZt ve^voSme of alr^at^Ihe r ^ vapours
ts found out from the tables. The ratio of tw^giTs
humidity i.c.,
M
volume of aifat'a g^vef actually present in a given
ENGINEERING PHYSICS
and ilf=mass of water vapours required to saturate the same
volume of air at the same temperature.
The amount of water vapours required to saturate a cubic metre
of air may also be determined experimentally. For this purpose the
above experiment is repeated after connecting a U-tube filled with
water at £>. The air passes through water before entering the tubes
containing CaClg and is completely saturated on passing through the
U-tube as the whole of the moisture is absorbed. By allowing a certain
volume of water to flow out of the aspirator, the quantity of water
vapours which saturates one cubic metre of air is calculated.
This method is quite a simple one but it is not so accurate as all
water vapours contained in air may not be absorbed. Moreover mass
of water vapours absorbed cannot be measured very accurately.
Usually, the apparatus is used for standardising other hygrometres.
Example 10. Two cubic Hires of moist air at SO'^C were drawn
ihrouyh a chemical hygrometer and 16'8 gm. of water was deposited in the
U-tubc. Find relative humidity if it is given that at 30^ C the mass of
water vapours required to saturate one cubic metre of air is 30. 1 gm.
Amount of water vapours in 2 cubic metres=16’8 gm.
1 6*8
Amount of water vapours in 1 cubic metre m— — - — = 8-4 gm.
Also amount of water vapours required to saturate i cubic metre
of air, i/=30-l gm.
8*4
• •
TTt 0*4
Relatove Humidity = ^ X100 = — ^XlOO =27 9%
(n) Dew poiDt hygrometers. These are the instruments which are
based upon the determination of relati?e humidity from the knowledge
of the dew point The dew point is the temperature at which the actual
vapour pressure at the room temperature becomes the saturated vapour
pressure.
The principle on which these work is that as the temperature
is lowered the surrounding atmosphere approaches the saturation
stage till at the dew point condensation starts and the surface of ‘{J®
instmment becomes dim. Thus the relative humidity is calculated by
using the relation.
Saturation vapo ur pressure (S.V.P.) at dew point.
^Saturation vapour pressure (S.V.P.) at room temperature
There are two kinds of dew point hygrometers which are named
as follows ;
{i) Daniel’s hygromoter.
{ii) Regnault's hygrometer.
We shall discuss here only Regnault’s hygrometer as it is more
accurate.
C HANGE OF STATE
219*
Regnault's dew point hygrometer. It consists of a wide glass
test tube A, the lower portion B of which is replaced by a thin polished
silver cap. The mouth of the tube is closed by a cork through which
are inserted, a sensitive thermometer Ti and a bent glass tube G
reaching almost to the bottom as shown in Fig. 9.20. A side tube CDE
the test tube A near the top and is connected to an aspirator P.
A similar tube B is fixed on the same stand and carries a thermometer
^ 2 - This tube is not connected to the tube A. It is
meant for the comparison of the brightness of the polished surfaces of
A and B when dew is formed.
To use the instrument the tube A is partly filled with ether so
that the lower end of the tube G and the bulb'of the thermometer Ty
are well within it. A current of air is drawn through the tube O
when the aspirator is worked. The air bubbles out through ether and
Fig. 9.20
surface of the tube B. Thl teXeJatwe^of thT th Pol'shed
noted. The working +v. ^perature ot the thermometer F, iq
of the thermometer fs Jga^n not^d wLn fheTw temperature-
I»l.t ..d “•"XdvS,u“X'X,”SX'S^
220
ENGINEERING PHYSICS
out from the tables, then
^ Saturated vapour pressure (S.V.P.) at dew point
‘"Saturated vapour prsssure (S.V.P.) at room temperature
The advantages of this apparatus are
(j) The rate of cooling can be controlled by adjusting the flow
of water from the aspirator.
(ii) The ether is in direct contact with the silver cap so that both
are at the same temperature.
(m) The ether is continuously agitated by the air bubbling
throughout its mass.
(iV) The first appearance of dew can be easily detected by
comparing the brightness of silver cap A with that of B.
(v) The observations are taken from a distance with the help of
a telescope so that breathing of the observer does not disturb the
humidity of air near the apparatus.
(iii) Wet and Dry Bulb Hygrometer. This instrument, as shown in
Fig. 9 21 is commonly used in meteorological
observatory. It depends upon the principle
that rate of evaporation depends upon dryness
of air. Smaller the humidity greater will be
the evaporation and so greater will be the fall
of temperature as evaporation causes cooling.
It consists of two thermometers fixed on a
stand side by side. One of them is simply ex-
posed to the atmosphere and is used for reading
temperature of the atmosphere. This thermc^
meter is marked as D and is known as dry bulb
thermometer. The other thermometer W has
its bulb wrapped in apiece of wet muslin
which is always kept moist by dipping its free
•ends into water contained in a small vessel B.
This is called wet bulb thermomer. Water
rises up in the muslim piece by the action of
capillarity and evaporates. Due to continuous
evaporation of water, cooling is caused as the
latent heat involved is being absorbed from
the bulb and mercury- This lowers the
temperature of the Wet-bulb-thermometer ^
The rate of evaporation and hence
of the temperature depends upon the state ot
humidity of air. Greater the relative humidity
lesser will be the the evaporation difference of
temperature in thermometer If thus ^When •'relative humidity
temperature between the two thermonae . no fall of temperature
-is 100 % then there will be no evaporation and so no iallot temp
Fig. 9 21
CHANGE OF STATE
22t
of thermometer W. In this case difference of temperature of two
thermometers will be zero. In all other cases temperature of JV will
always be less than the temperature of D. Hence the difference
between the temperatures of the two thermometers is a measure of the
Empirical tables have been prepared from which dew point and
relative humidity can be calculated if the lowest wet bulb reading and
constant dr}’ bulb readings are known.
(tv) Hair hygrometer. It has been found by experiment that human
hair expands with increasing relative humidity. One form of hair
hygrometer is shown in Fig. 9.22. A long
strand of human hair carefully cleaned from
all oily matter is fixed at one end A and
passes over three small pullys h.c and d.
The hair is then wound a few turns over
a revolving small drum e and is finally
fastened to the end of a spring s at /. The
spring .S' acting on the drum in the opposite
direction keeps the hair taut and the
changes in the length of hair cause the
movement of the pointer P over a scale
which is graduated directly in terms of the
relative humidity.
(») Under normal workingconditions,
and when subjected to ordinary variations
of humidity and temperature, the hair hygrometer tends to eive
high readings with lapse of time. This is probably due to the fact that
the hairs are always under tension and consequently permanent
(n) The most reliable inetruments are those in which the tension
force IS small and the hair is safegaurded from any undue strain.
{in) The calibration of the instrument should be checked at
frequent intervals under normal working conditions.
(tv) It is very important to see that the hair should be free from
grease and before it is inserted in the instrument it must be washed
Lbsornf **^^”^f otherwise presence of grease will interfere with the
un^Sk! ^^^der the readings of the inrtrument
If suitable precautions are taken the error should not be more
cSrd''stor^at'^‘'"!i usually sufficient for practical working in
the hygrometers are often used there. They Dosfe<;«i
be reading instruments. Further ^thev can
be employed at a temperature below 0®C where the wet anH Ht-
liygrometer present difficulties. ^
9.26.
efficiency
Importance
of a man
of Hygrometry. The comfort and the
IS very mucli affected by the
working
seasonal
222
ENGINEERING PHYSICS
variations of temperature, humidity etc. In winter when
the temperature is too low, we feel chilly and want to heat our
rooms. In summer, we feel hot and wish to cool the rooms. Besides
temperature, the humidity of the air also has a great effect on our
physical comfort and our health. On a rainy day in summer, we
feel far more uncomfortable than on a dry summer day, though the
temperature in the later case may be slightly higher. Thp. reason is
that, on a warm rainy day, the humidity of the air is high and the moisture
given out by our bodies does not readily evaporate so that we feel
uncomfortably damp. But on a fairly dry day even when the temperature
is higher, the moisture from o%ir body soon evaporates with consequent
■cooling, and we feel more at ease. Similarly, if, in winter the humidity
of the air becomes too low, the membranes of our nose and throat
are rendered dry and cause considerable discomfort on such a dry
day, the skin on the face often cracks producing an itching
sensation. Relative humidity of about 50%, is found to
be ideal for health. The above facts can be rewritten for complete
explanation when the relative humidity rises above 50%, the rate
of evaporation from our bodies is very much reduced and we feel
oppressive and the rooms appear to the stuffy. When the relative
humidity falls below 50% the air becomes dry and evaporation from
our bodies is rapid with the result that our throat, nose etc feel
dry and thus produce a feeling of irritation.
Hence there is desirability of controlling the humidity in the
offices and homes besides the controlling of temperature. In addition
to the controlling of temperature and humidity, we also make
arrangements for proper circulation of air and to remove dust
particles and undesirable odour to make it more comfortable for
human beings.
Humidity affects weather, hence regular observations of the
changes in the atmospheric humidity help us in predicting the
weather which is useful for meteorologists.
The successful working of cold storages for fruit, meat etc often
depends upon tire degree of control of water vapours present in tne
enclosure.
The control of humidity is also very important in textile industry
specially where cotton and woolen goods are manufactured. Humiaity
also plays an important role in the artiffcial seasoning of
manufacture of tobacco. It also plays an equally
in the air-conditing of building both in summer and winter seasons.
The process of regulating the temperature, humidity, purity and
circulation of air is known as Air-conditioning.
EXPECTED QUESTIONS
1. What is meant by 'fusion' and ‘point of fu.sion
difference between crystalline and amorphous substances giving
? Explain the
a few examples
in each case. . . .^.^-cnre
2. Explain what is meant by regelation ? _How will ^
effect the melting point of (i) ice (**) wax and ^ STon^solidification.
importance of the substances (like antimony etc.) which expand on sohdihcatio
CHANGE OF STATE
1 ?L define the freezing point of a solution ? Is it higher or
lower than that of pure solvent > Give a quantitative relation between depress-
ion Of freezing point and the concentration of the solute What is molar
concentration ? moiar
4. Explain fully the following : —
(t) regelation (ti) snow balls («i) skating.
5- Define the following : —
(<) I.e Chatelier’s principle
(fi) Eutectic alloy
(m) Eutectic mixture
(to) Eutectic proportion
(»’) Eutectic temperature.
6. Write short hotes on : —
(*) Triple point
{{>:) Dew point
(•♦0 Kefrigeration
(/y) Steam line
(y) Hoar frost line
(ot) Ice line
ivt ) Sublimation and
(viii) Kefrigerant.
7.
(*■)
(-•<)
(*•)
Hi)
What is meant by the terms
Saturated and
Unsatuted vapours.
Discuss the relation between : —
pressure and temperature and
pressure and volume in the case of vapour.
compr!km^“ saturation vapour pressure ? What is the effect of
(0 4 saturated vapour and
(u) unsaturated vapour ?
How is the maximum vapour pressure of a liquid related to its boUing
9.
point ?
9.
point
Delhi. Explain^it‘“u^ 5 !'““^“‘““““"‘>^‘=‘>“'"‘ than au equally hot day
CHAPTER X
EQUATION OF STATE
10.1. Equation of State. The relation between pressure, volume and
temperature of a substance in any state of its aggregation is called
the equation of state.
For a perfect gas which obeys both Boyle's and Charle’s Laws the
equation of a state is expressed as,
PV=RT.
10 2 ^Andrew’s Experiment. An ideally perfect gas is that which
strictly obeys gas laws^ 1..., Boyle’s Law and Charles’ Law. No gas is
perfect in the true sense of the word. All gases show some deviation
from the eas laws particulary at low temperatures and high Pressures.
T^deviaW gas iLs is also intimately connected w^h the
the deviations are more marked
carbon dioxide.
Andrew’s Experiment. other^ paru“^Cng of
of whicli was a graduated capillary tube tne
wider bore, and passed carbon le<j and the lower
remove all traces of air. ®°th ®n ’ surface ’ in a container,
was subsequently reopened under a mer y t
which could be partially ° d ^by a pellet of mercury,
adjusted to tlie required value und trapped by P ; above
The tube A was then mounted with anotl er tube ^ n b cylinders
mercury thread, in a vessel consisting of^two^^r^on„^ copp^
filled with water ^ud hav. ^ t connected by a cross
bottom as shown in Fig. 10. l. i^oui
tube as shown thus be produced and
224
EQUATION OF STATE
22S
ressing the water by means of either or both screw plungers. Andrews,
however never went beyond I 08 atmospheres in his o^vn experiment.
tig. lU.l
rfefermtnc(f from the volume occupied by air for
(Ae rehition behoeen pressure, volume and temperature is accurately
The projecting parts of the capillary tubes were enclosed in
aesirea temperature between o and lOO'^C and the other amnnH n K
represented* by a'itf* volumn^teseTlrthrtf 8^®'
approximate obedience to Boyle’s W where ‘"'teases in
gas behaves like a saturate7vtpo^ 3 i[n " ‘he
the volume diminishing rapidly^and the press^urJ* tp with
at the value reached at 6: This proffres?iv^ remmmng constant
conWs along the practically horizln?2 pirt ^rofth
brings about a negligibly small decrease"' in
ENGINEERING PHYSICS
almost incompressible. This is shown by the part c d which is a
straight line almost parallel to the y-axis.
Fig. 10.2
(ii) at 21.5®C. The general shape of isothermal for a temperature
21.5°C is again the same. The gas obeys Boyle’s law in portion AB
which is greater than a b. From B to C the gas behaves as saturated
vapour and is in contact with its own liquid. The volume decreases
and pressure remains constant. From C to D it is in the liquid state
and increase of pressure produces no decrease in volume. The second
isothermal differs from the first in the following respects : —
(i) The volume occupied by a given mass of the saturated
vapour when condensation starts is smaller : the volume of the
liquid when condensation is complete is larger than that for the first
case ; and {Hi) the horizonal part BC is smaller than the part be. This
shows that with rise of temperature the density of saturated vapour
increases and that of the liquid decreases.
(it) at 31.1°(7. For the isothermal at 3T1®C, the horizontal
portion of the curve has almost disappeared. The curve shows that
from J to K and K to L. Boyle’s law is obeyed. In fact, the highest
temperature at which liquefaction occurs and the isothermal shows a
horizontal portion is 30-90®C.
This temperature is known as critical temperature and the
isothermal corresponding to it is called critical isothermal.
(if) At tempperatures above 31 -l°C.The isothermals for temperatures
35 ' 5 ®C’ and 48'1'’C much above the critical temperature do not show
any horizontal portion and obey Boyle’s law from one end to the
other. The isothermal for 48'I®C' is similar to the isothermal for air
shown separately in the same figure.
Discussion- (i) The dotted curve joining the e.xtrernities of
the horizontal part of these isothermals is called the horde.r curve.
Its ape.x J/ is called the critical point and it lies on the critical iso-
thermal. The border curve clearly indicates the different states ot a
substance. To its right the substance is unsa Curated vapour, witmn
it a mixture of saturated vapour and liquid and to its left entire y
in the liquid state.
EQUATION OF STATE
227
{it) For every gas there is a critical temperature below which a
gas must be cooled m order to liquefy it by means of pressure alone.
At this temperature the volume and density of the saturated vapour
become equal to the volume and density of the liquid. It is not
possible to distinguish the vapour (or gas) from the liquid.
The gas behaves almost as a perfect gas above the critical tem-
perature but shows large deviations from Boyle's law at tempera-
tures below the critical value. The pressure necessary to liquefy a
gas at the cnUcal temperature is called the critical pressure and the
volume occupied hy a unit mass of the gas at critical temperature is
called critical volume.
[ill) There exists a continuity of state between the liquid and
the gaseous phases. If we start with a given mass of the substance
an the state represented by the point P. heat it at constant
volume until the point Q is reached, and then cool it at constant
pressure so that the point R is reached, then the substance changes
from gaseous to liquid form without any discontinuity occurring. ^
***!"*** may be regarded as distinct
stages of a senes of continuous physical changes. In other words a gas
and a liquid are widely separated forms of the same
matter, which can be made to pass from one to the other by a series
of changes so graded that no breach of continuity occurs^ Thurat
high temperatures and low pressure the occurs, inusat
tur^ high pressure the substance entirely becoS
deviation is small but^at high pressureT^and^T^^ pressures the
deviations are very large In the derivaf temperatures the
basis of kinetic theoryTi gases
which do not hold true in fhe case of real gaTesl'^^hesfa^re
w The molecules are mere mass points having no finite volume
TuollciLs;"""^ or repulsion betweenTwo
Joule -Thomson on the liquefaction of “f
exists a force of attraction between tl shown that there
this force of attraction is much less thin a gas. Though
fh‘ negligible Van der^W,’"! liquids
the perfect gas equation® /K=itT based ^ it rejected
assumptions of the kinetic theorv vir ee ^ simplifying
zero force of attraction between th4m 'tf molecules and
equation of state for actual gases takW ® modified
6 s, taking into account both
228
engineering physics
molecular size and inter molecular attraction. His equation has not
only the merit of simplicity but its general applicability over a wide
range of pressures and to larger number of gases.
Correction for the finite size of the molecules. Let CDEF represent
a closed space of volume V containing the gas, in which the molecules
of the gas are assumed to move. For our present purpose, the mole-
cules are considered as hard elastic spheres with a finite radius.
Consider a molecule A moving parallel to CD and suppose it strikes
the face of the vessel DE at B and comes back. Now if CD=l, the
distance travelled by the molecules is not 2l but
2 [l—d] where d is the diameter of the molecule.
If we consider the motion of all the molecules
in this way, we find tha t the effective, y olume in
which the* molecules move is less than the total
volume T' of the enclosure. When the pressure
is increased, there is a decrease not in the whole
volume but a part of it i.e.. (F— 6) where b is the
volunse of the portion which can not be compress-
pH This constant b is four ti mes the total-
volume of the molecules aH^^^noT^ual to the Fig. 19.3.
volume occupied by the molecules, because,
(a) When all the molecules are moving about, they obstruct each
other’s motion much more than if some of them were at rest.
lb) Each molecule is surrounded by its sphere of influence the
radius of which is greater than the radius of the molecules w ithin
which no other molecules can penetr ate.
Thus 6 is the reduction in the effective volume of the gas and
is called as co-volume, the ideal 'gas equation, as corrected for
molecular size, thus becomes,
P {V~b)=BT
or
It is clear from the equation that the effect of the co-volume is^
to increase the observed pre^ure.
Correction for inter-molecular attraction.
forces in a gas are probably of the same nature as
rise to the phenomenon of surface tension in liquids T
Attract one'^another with a force which vanes inverse^ as some^pow-
of the distance between them, so that the attraction apF
only for small distances and is negligible for large ones.
Considering a molecule^ samrfree and the'neltrce
other molecules in directioris _ohitTnea^^^^ wall of the
on it is zero. But when it finds itself -or to unbalanced
vessel, it will experience a pull towards the in terior d ue
229
EQUATION OF STATE
molecular forces. This pull directed towards the interior and normal to
the wall increases from a minimum at a distance
equal to the radius of the sphere of influe nce to a
maximum just when the molecule is about to
impinge on the wall. Hence the molecule will strike
the wall with a velocity less than that with which,
it would have struck in the absence of inter-
molecular attraction. This. is true of every
molecule striking the walls of the vessel. The
velocity and hence the momentum with which the
molecules strike the walls is, therefore, reduced due to the inter
-molecular attractions. Naturally when the momentum becomes less
the pressure decreases.
The magnitude of this decrease in pressure may be estimated as
follows :■ — ' ■
i
Fig. 10.4.
It is due to simple reason that the decrease in pressure is pro-
portional to
(t) the number of attracting molecules per unit volume, and
(it) the number of attracting molecules striking a unit area of the
walls of the containing vessel per unit time.
Both these factors are proportional to the number of molecules in
unit volume of the gas or to the density of the gas.
Decrease in pressure px (density of the gas)*^
1
• 4
cc
V*
or
where a is the constant of proportionality.
Hence the real pressure ^Observed pressure-f-decrease in
pressure
and this value of real pressure should be substituted in the perfect gas
equation.
Now, substituting (V-b) tor V and for P, we have
( ) (V-b)=RT
This equation is known as Van-der Waal’s equation of state.
Compari^n Witt experimental facts, Van-der- Waals' eouati
IS not perfect. It is only reaching perfection. We will t^st t
theortical results obtained from Van-der-Waals ' equation by dra
230
ENGINEER(NG PHYSCrs
o^Andre". With the
perimental curves ^egiVe^n In F^g/lO^^ in Fig. 1 0. 5 whereas the ex-
closely with the experimentlTone^s*^^'"^^ temperature agree
similar to°hrpa\t^AB ^he part A'B' is
the portion CD. ^ portion C'D' is also similar to
portioi Bcf ^ whTreaf etn** ' with the experimental
fheoretical .^urve shows a maximu^^^a™ X and
a maximum at JL and a minimum at Y.
Fig. 10.5
This devliation has been explained by Thomson considering the
phenomenon of super-saturation.
If liquefaction had started at B' the portion B'X would have
been absent. It, therefore, represents the unstable state of super-
saturated vapour. It is possible under certain conditions to compress a
vapour beyond the saturation stage without liquefying it. The portion
C'Y represents the unstable state called the superheated liquid. It is
sometimes possible to heat a liquid above its boiling point without f allowing
the liquid to boil and change into vapour form.
The portion XY represents an anomalous behaviour, tiuit ths
volume decreases as the pressure is decreased. It cannot he realised in
practice.
231
EQUATING OF STATE
10.5. Van-der Waals eqnation and critical constants. Van-der Waals
equation may be written in the form
{p+
where P. V and T are the pressure, volume and absolute
of the gas respectively and a, b and R are constants,
ing the above equation we get,
temperature
On expand
PV-Pb-{- y
Multiplying by T®, we have
py3^PbV^-haV-ab=V^RT
Rearranging the terms, we get
PV^—V^Pb+RT) +aV~ab=0
Dividing by P, we get
This is a cubic equation in V and has iAree roots in general,
there are three values of V for a given value of P and T. At the critical
point all the three roots will be equal since a gas has only one
value of volume, pressure and temperature corresponding to it.
Let X be the value of this critical volume, then
V=x or r — XasO
(7— «)«=0
or 73— 7». 3 x+7. 3x^—3^=0 ...(h)
Comparing the co-efi&cients of 7*, 7, and the ‘constant term
in (*) and (it), we have
...(»«)
II
232
ENGINEERING PHYSICS
or
Dividing {v) by (iv), we get
3 P
or x=36
Critical volume T’c =36
Substituting the value of a: in equation (iv), we have
3X962=-^
or
a
Critical pressure Pc~
276*
a
iW
Substituting the value oix and P in (m), we have
RT
3x36=6H X276*
a
ZaT^nbRT
and Fo=36 /
From the above values, it is seen that
RT.
8
This ratio is called critical coefficent though its theortical value
is its actual value found experimentally is much more than this for
different gases.
Example 1. Calculate the value of critical temperature for CO^ for
which a=0-00874 atmosphere and h=0'0023 c.c.
We known that
Tr.=
Za
21 Rh
The value, of a and 6 given are in atmospheres and in c.c.
respectively. To find the value of critical temperature let us first
EQUATION OF STATE
233
calculate the value of J? at a pressure of ! atmosphere for a volume of
2 c.c. at 0®(7 or 273®^ from Van-der Waal’s equation.
( ^+-fJ (^-b)=RT
Substituting the values we get,
f 1 , 0 00874\ .
) n — 0-0023)=/? X273
or
R=
Now
8a
100875X 27 2
273
8X0-00874X273
nRb 27X1*00874 X 0*9977 X 0*0023
= 306®K or 33®C
10.6. Defects in Van-der Waal’s equation.
(2) It does not tell us when the condensation begins.
. values of a and 6 do not remain constant at all
temperatures,
(3) If we calculate the critical constants we find that
RTc 8
PcVc ~ 3
for all gases but actually volume of varies with the gas and has.
* c
an average value of 3*77.
expe[?Ln1X theoreticaUy, though
relation LtwtrtteTes^^^^^^^ An ^.^ich expresses the
the equadon of a s^e ctra^tr^
pressure, volume and temoerature S we express
respective critical values and^subsStut^^fwff® as fractions of their
Waal’s equation, we obtain a rednrAH for and T in Van-der
same for all gases. This principle is state, which is the
of a Sri ?racKf t~s%c^W:crmral L“s^, w^eTJe'
Pc
y « T
■■K, and
Tc y*
Vc
%
r •
234
engineering physics
so that P=aPc. V—^Vc and T yTf.
where a, p and y represent fractions and are
(representative) pressure, volume and temperature resp V
together referreo as reduced variables of state.
Substituting the values of P. V and T in the equation
^ )=BT, we get
Now
Pc =
276*
, 8a
. r,=36and
(«+-rXp- 0 =T>'-
This is the reduced equation, of ^
fj.T ffsr, Slip's t
respectively. Then clearly
and
V ra ' '
Obiviously therefore if «,=«. and -e shall have n=n_
t0.s. Liquefaction or gases. a
^i^^iirgl^. ftT ihe “So^dfwere liquefied
of sinrple -ainS
T npral there are two methods by which second method
T.A"rs'f ' h- •»■ ■■”“
^ppl^lng a sStly high pressure.
EQUATION OF STATE
235-
r
The three main methods by which gases can be cooled are : —
!• Cascade process, which utilizes a series of liquids with
successively lower boiling points to reach the low temperature in
stages.
2. Regeoerative Joule — Thomson, process, based on the Joule
inomson effect and regenerative cooling.
3. Adiabatic expansion prorcess. based on the cooling produced'
when a gas expands adiabatically doing external work.
f r Pictet or Cascade process for liquefying oxygen. If we wish
10 iiquefy oxyen, we must in some way cool the gas below its critical
temperature apart from any increase in pressure. In 1878 Pictet
wording independently brought about the liquefaction of oxvgen.Pictet
made use of the cooling effect of a liquid which was rapidly evaporat-
ed. His method is often known as. Cascade Process and it may be
aescnbed as a number of compression machines in series, each
machme reducing the temperature of the liquid to a lower value than
^ sas like methyl chloride (CB^ Cl) is
liquehed and then allowed to evaporate under reduced pressure. The
ecnyiene {C.H^) by passing it through a spiral coil immered in liquid
to
the
is-
236 ENGINEERING PHYSICS
passed through ethylene. Oxygen liquefies at a pressure of about
20 atmospheres, when its temperature falls below— 118°C. Oxygen
therefore, must be cooled to a temperature below this particular
temperature if it is to be liquefied. About the same time, Earner
leigh Onnes also employed the cascade process for liquefaction ^ of
oxygen, using methyl chloride and ethylene for his two working
substances. His apparatus is shown diagramatically in Fig. 10.6
It consists of three compressor machines A, B and C working in
series by means of three pumps P, Q and R. Machine A is filled with
methyl chloride and surrounded by cold water. Sii^e
temperature of methyl chloride is I43®C, it can be liquefied at the
room temperature with a pressure of few atmospheres.
Compressed methyl chloride gas from pump P is passed
-the spiral tube (shown straight in the figure), inside the compressor
machine A. while cold water is circulated through the inner l^c^t ot
A. Liquid methyl chloride thus obtained IS made to circulate tn
jacket of the compressor machine B of the second '"''V .
wnnected to the suction side of P. This enables the liquid methyj
chloride, which normally boils at -24''C, to eyapomte
pressure and produce a fall of temi^rature to about 9 |
impressed Ethylene from pump Q is Passed tempe-
tube inside the compressor machine B. Since it has a
Tature of 9-5”(7, it is readily liquefied in B
— 90‘’C by the continuous circulation methyl • j^^t^of
it Liquid ethylene formed in B is made to flow round the jacket oi
iU,;: Ais's.
TC';..uK.»a b, b.n,«
oxygen under reduced pressure is —21 8 C.
This method may be continued in order it Is not
err; (l--. — ™-
Kelvin (Thomson) effect. educed when
Toule performed experiments to find cooii g^^P
compressed ail in one v^sseljas^ alio wed P ■ in^i-
evacuated vessel, throug temperature was observe •
cated that in the 'Y^ole no chang ^curate results as the *ieat
experiments were , /'omSared to the heat capacity of the
capacity of the gas was small as compar
K:alorimeter used by mm.
EQUATrON OF STATE
237
Joule and Lord Kelvin (Thomson) later performed more accurate
experiments and found that the cooling effect was observed in all
gases except hydrogen. The cooling produced when a gas is allowed to
expand through a narrow orifice is called Joule-Kelvin effect and it was
proved by this principle that a gas is cooled when forced through a
narrow orifice provided the temperature at which the process takes
place is below the temperature of inversion for this gas {this timpc^
rature being ahrays higher than the critical temperature).
If air is being liquefied, it is first freed from carbon dioxide and
water vapour by being passed through caustic soda, since during the
process these would solidify and check the pipes. It is then comp-
ressed to about 200 atmospheres by the pump P as shown in the
Fig. 10.7 and the heat of compression is removed in the coil C
immersed in the water bath which is at room temperature. After this.
r ig.
ft v;« /
the gas passes through a spiral in the heat exchan^^i-r
through nozzle N, the aperture ot which can be controlled
^ide the exchanger. This produce- coolinrand the ‘ “ °*'‘-
through the outer tube of the spiral to the^umo P ^h up
progressively cooled, and cools the air approachinc the ^
the temperature is reduced below the critf^al value^wW ?h
IS sufficient to liquefy the air which then cnlleef • Ou pressure
Heat gain from outside is redu“y the Wv
material / which surrounds the heat exchanger^ insulating
inversion.Temperature *^6*^ mucV‘2Sf^^^^ Jh their
ratures and there is no dfficX in rf^Mn. tL
the Linde process will operate. ^ '^®»<=n>ng the temperature at which
ENGiNEEKiNG PHYSICS
2£8
3* Claude Process* The air liquefier depending on the Joule Thomson
.effect is a some what inefficient arrangement thermodynamically
and in 1902 Claude devised an apparatus in which the compressed
gas is allowed to expand adiabatically in an expansion engine and
consequently does external work In this case there is cooling due
to the external work done in addition to the cooling due to the
Joule-Thorason effect. Further more the external work can be employed
in helping to drive the compression pump.
As shown in Fig. 10.8 the compressed gas, purified and precooied
-to room temperature, passes down the tube A and part of it is allowed
^o expand in B driving the piston and so doing work thereby cooling
to a low temperature. This cooled gas at low pressure is sent back
to the condenser C where it cools the down coming high presswe
gas on leaving C. the low pressure gas is conveyed back to the
Fig. 10.8
compessor via the jacket where it cools the incoming gas also. In
ttiT way! the gas is progressively cooled to lower and lower temp-
raJure until its temperature falls to a value below its critical
temperature. When it starts liquefying under the incoming pressure.
The^ resulting liquid can then be drawn off by means of a tap T
from the botton of the condenser C.
-It is difficult to find suitable lubricants for
neratures although Claude » found petroleum ether to he smtabl
for his work. In practice the method is not nearly as e
would be theoratically predicted and most commencal hqu • ii^
The Linde process since ft is only slightly less efficient, is mechanically
simpler and requires no lubricants.
10 9 Rcfrigeratiog machines. These are the Pf the
proLctioro® artificial cold required to maintain a chamber for
EQUATION OF STATE
239
storage of parishable food or fruit at a low temperature or to keep a
brine^bath well below 0°C for manufacture of ice.
The basic principle involved in these 'tnnchines is the cold produced
by evaporation of a liquid under reduced pressure, often in conjuclion
with the cooling effect produced when a gas expands adiabatically as
already mentioned.
The liquids which are generally employed are : —
{i) Ammonia (It is most commonly used in large referigerating
plants as in ice factory).
(n) Sulphur dioxide (^SOo).
{Hi) Freon or dichloro-difluoro-methane {C Ff)
The liquids (it) and (Hi) are employed in small referigerators
rneant for domestic purposes. Sulphur dioxide can be more easily
liquefied than ammonia or Freon, so that a less stout compressor
and condensor can be used, but for producing the same cooling
effect as with ammonia, the compressor has to be made large.
Two types of refrigerating machines are used in practice. These
are (i) compression type (ii) Absorption type.
The essential difference between the two consists in the manner
of compressing the low pressure vapour. In the first type of machine
»•€., m the compression type a motor compressor is used while in the
absorption type a dilute solution of the refrigerant at room
temperature is used to disolve the low pressure vapour and the
o^centrated solution is heated to expel the vapour at high pressure.
Ine compression type is more efficient, requires less initial cost and
ence is used much more than the absorption type. The compression
ype referigerating machine is called Frigidaire or vapour com-
pression refrigerator.
The principle of action of this type of machine is shown in Fig. 10.9.
NKiH
PRESSURE
StDE
LOW
PRESSURE
SIDE
WATER
WATER
PROM COLO
STORAGE
CHAMBER
BMWE
SOLUTION
TO COLD
storage
CHAAtBER
TNROTTU VALVE
The piston
Fig. 10.9
A is driven by an electric motor and is connected through
240
ENGINEERING PHYSICS
a valve E to a. spiral (evaporator) in the cold bath B and through
valve F to a spiral or condenser in hot bath C. The spirals are
connected at the bottom by a tube and throttle valve V and contain
suitable chemical such as ammonia or sulphur dioxide. In modern
machine dichloro difluoro-methane (Freon) is often used.
When the piston moves upwards, the pressure in the cylinder falls
below the pressure in the evaporator and as a result the valve E opens
and the low pressure vapour is sucked into the pump through the suction
pipe During the downward storke of the piston. the vapour is compress-
^ which closes the valve E and opens the valve F.and the compressed
vapour is delivered to the condenser through the discharge pipe.
The condensor C is cooled by circulating cold water and the vapour
liquefies due to high pressure and lowering of temperature. The
liquid referigerant passes through the expansion or regulator va ve
r which reduces the pressure of the liquid from the high value
nrevailing in the condenser to the low value in the evaporater. Due to
low pressure the liquid evaporates in B. extracting its latent heat
from the cold storage space or brine water surrounding B. The space
or brine water is consequently cooled. The low preseure vapour is
sucked in again by the compressor and the cycle of operation
is maintained,
10.10. Domestic Refrigerator.The principle of the working of a domestic
70 MAINS
/iECULATINC Ml V£
FR£EZ/N6
CHAMBER
Fig. 10.10
reirigerator is similar to the ammonia ice plant. It consists of a
EQUATION OF STATE
241
small double walled cabinet or box, the space between the walls is
packed with a heat insulating material like glass wool. The com-
pressor and the condensing coils are placed at the bottom of the
cabinet. The evaporating coils and the regulating valve are fitted in
a small chamber near the top. The gas used is freon, i.e., di-chlorodi-
fluoromethane.When a low pressure is created in the evaporating coils
freon vaporises producing cooling. The temperature in the chamber
falls below zero. This chamber is known as the freezing chamber and
is used for freezing wate*- or milk.The cooled gases leaving the freezing
chamber are circulated in the rest of the cabinet thereby producing
a low temperature of about S^C. The cooling of gas in the condenser
is done by circulation of air with the help of fan. A thermostat is
provided in all refrigerators. This makes the working of the electric
motor automatic. The motor starts as soon as the temperature
rises above 5^(7 and stops when the temperature falls below 5 ®C. A
refrigerator is used for storing food, blood vaccine, fruits, etc., which
keep well when cold but deteriorate at ordinary room temperature.
Expected Questions
an account of the work of Andrews on the relation between th^
temperature of a gas. Explain with the help of diagrams the
terms critical temperature and critical pressure. ® ®
2. Describe a method for determining the critical temperature.
criticaf’nniii^*' properties of a fluid in the neighbourhood of the
‘ critical constants of a fluid and describe how they are
derivation of Vander Waal's Equation of state. Discuss the
efluatifn equation of state of a gas from Van-der Waal's
they will have th^same
7. Distinguish between a vapour and a gas.
Descri^be of gases.
9. Describe a method for producing liquid oxygen for industrial purposes.
chapter XI
THERMAL CONDUCTIVITY
Heat travels from a body at a higher temperature to a body
at a lower temperature. There are three distinct processes by which
heat may travel from one place to another.
I. Conduction 2. Convection and 3. Radiation
II. 1. Conduction. We know that heat is a form of ener^ possessed
bv a body by virtue of the vibratory motion of its particles.This concept
at once helps us to understand the conduction of heat along a certain
materTal When one end of an iron rod is placed in fire, the nrolecules
near this end receive some heat energy from the fire ^nd^g'n to
vibrate more vigorously. These molecules collide against and agitate
t £
,h, ..a
are generally heated up by this process.
conductivities.
Those substances which allow heat to wood,
great difficulty are called bad bad conductors. As a rule
» Vd'ortsfe,.,. b.. M.,-, .b- »>r
exception.
242
therhal conductivity
243
11 . 2 . Thermal CondactiWty. Consider a cube of some material whose
opposite parallel faces are maintained
at different temperatures. The quantity
of heat Q that flows from the face at
a higher temp,, 6 ^ to the opposite face
at a lower temperature 62 , is found to
depends upon the following factors.
(») It is directly proportional
to A, the area of the face.
(n) It is directly proportional to
the temperature difference ( 61 — 62 )
{til) It is directly proportional to
the time t in seconds.
Fig 11.1.
(iv) It is inversely proportional to the distance between the
faces i.e., distance x.
(u) It depends upon the nature of the material.
Then «a
X
two
or
X
Where ^ is a constant depending on the nature of the material of
the cube and is known as its thermal conductivity or coefficient of
thermal conductivity or co-efiicient of heat conductioo.
If in equation (t) above.
A=l sq. cm., 02 ) = 1'>(7
t=l sec, X — 1 cm., then
Q=K,
a
<ube of the material having area of ZZnle “
between the faces one cm, when thi ffu-p? distance
temperature difference of 1°C, ' maintained at
Or
<>neIeZdTrZ;fLTfZeZaZ:^^^^^ •’J ^^oHes that flou,
snaintained at a temperature difference of rC, ^PPosite faces
Re-arranging equation (i) we have
Qx
'8 in
are
K=
A{0^-6^)t
244
ENGINEERING PHYSICS
(a) C.O.S. units
Q (cal) X (cm)
A{sq. cm. or cm.*) (sec)
cal . 1 1 t
~ vm — r-T- . oTi =cal cm“^ °C-^ sec-'
A (Wj — Og)^ cm. C sec
i.e. the units of K are cal/cmrC/sec.
( 6 ) 31 . K. S. units. In this system metre is denoted by m and
kilocalorie by K~cal.
K=
Q{K ~c&\) X {m)
A{m^) X (sec)
Qr
X-cal
^{^1— dg) ^ ' m ®C sec
-=/f— cal m-' sec-'
i.e., the units of K are K— cal/m/®C/sec.
11.3. Effect of temperature The thermal conductivities of practically
all materials depend upon the temperature of the material. For some
materials the thermal conductivity increases as the temperature of
the material rises, for other it decreases. This variation is approxi-
mately linear over a consideaable range of temperature for iriost
materials. Thus, if the thermal conductivity is plotted against
temperature a straight line is obtained as shown.
Fig. 11.2
11.4. Temperature gradient. We have seen that there is a temperature-
fall of (^1 — ^2) over a distance x. The quantity 's the rate at
which temperature changes with distance in the direction in which heat
flows. It is called the temperature gradient.
If the distance between the two faces is very small and is denoted)
THERMAL CONDUCTIVITY
245
by dx and the difference of temperature by dB, then the limiting value
, ^ 1—^2 dB ,
temperature gradient.
Hence equation (t) can be written as
Q=KA^ t
dx
Since temperature decreases as x increases, hence sometimes a
negat^ive sign IS put on the right hand side of the above equation
Which becomes. ^
Q^-KA^~t
dx
distance temperature decreases as the
instance x from the hot fac« increases.
the body is said to be unsl..ady if
safd t^hf , j f"y P'»“‘ >" ‘be body varies with time and it is
said to be 3teady if the temperature at every point in the body remains
constant. Hence with unsteady conduction the lempetraure gradient —
at each point in the body also varies with the tim^ \X 7 ;.k *
conduction the temperature g'radient at eacrpo'i^nimatus ^ns'’tatt“^^
insulatL^T^’l®’ '"ben steam is first admitted to a pipe covered with
point?is Jit^'b^^ ^ two
difference betw^i^n two points
•« Resistance betweenThese poin^
But i2=^L=_L
a
acr
Where
^=“^=electric conductivity
2
P
i=a^
ao*
I±=length of the wire
a-area of cross-section of the wire
(«•)
246
ENGINEERING PHYSICS
(ft) the rate of flow of heat between two point is given by
=KA — ^ cal/sec
where
Q
t
or
X
KA
Comparing (it) and (tit) we see that temperature difference (tfi—
corresponds to potential difference V and corresponds to electrical
resistance and may be called as thermal resistance.
heat cuiTent=
Temperature difference
Thermal resistance
In an electrical circuit if there are two or three resistances in
series, then the current is given by
V V
• • »
® 3^3
acr
Similarly if heat conduction takes place through two or three
plates of different materials then equation for heat current becomes.
e.-e
a
+
X
s
£*_+•
^ AK
vartuf ^artfofthe ^ody rJLa^^rtant «
state is reached, the rate at whidi a b^y 6^= ..^^acity of
only on the thermal conductivi y Quantity called thermo-
the body. We therefore, introduce another quantity
metric condnctirity or thermal diffnsiTity (A).
THERMAL CONDUCTIVITY
247
Thermometric Condactivity or (diffusivity h) ia defined os
the ratio of the coefiicient of thermal conductivity to the thermal capacity
per unit volume of the material.
If S is the specific heat and p the density (mass/iinit volume).
then Thermal capacity per unit volume x P
Thermal conductivity K
~ Specific heat x density Sxp
Example !• A boiler ia made of iron plates 1‘2 cm. thick. If the
temperature of the outer surface be 120°C and tha.t of the inner 100**C,
calculate the maaa of water evaporated per hour, assuming the area of the
heating surface to be 50,000 sq. cm., K for iron=^0’2 C.O.S. units.
Heat transferred per hour cals
X
_ 0-2X 5 0,000 (120 — 100) X60X60
1*2
=-6X 10® cals.
Heat required to evaporate one gra. of water at I00'’C7 is equal
to its latent heat =540 cals.
mass of water evaporated per hour
6X10®
540
= 1-11 xio® gms.
j f^ample 2. A slab of thermal insulator is 100 cm* in cross-section
^ cm. thxck. Its thermal conductivity w 2 X 70-^ cal. cm'* sec-^
If the temperature difference between the opposite faces is lOO^C how
nuzny calories flow through the slab per day ?
X
Substituting the values of
we get
various quantities in the formula,
2X10-®X100X 1 00 X 24 X 3 600
2
= 86400 cals.
=86*4 K. cal.
could 6e raised per hour at atmosviUtic
248
ENGINEERING PHYSICS
and tin are 0'9 and 0'15 c.g.8» units respectively. Latent heat of steam at
normal pressure is 540 calfgw.
Heat is conducted through the copper plate and the coating
of tin connected in series
d$ Add
Heat transferred Q --^ — X
Zx
KA
Sx
K
xt
100xf700-100)x60X60 ^ Cals,
■ ■
0-24 , 0*02
Heat required to convert 1 gm of water at lOO^C into steam
= 540 Cals.
Maximum mass of water evaporated per hour
— ; =10® gms.
Example 4. A slab consists of two parallel layers of
materials 4 cms and 2 cms thick and of thermal conductivity of O‘04and
0-36 c.g.s, units respectively. If the opposite faces of the slab are at lUU O
and O^C, calculate the temperature of the interface.
Let the temperature of the common surface be Q°C.
KiA {e^—e)t K^A
a:i arg
0-54 Xv4f 100 — 9)t 0’36 A ( 6 — 0) t
2 ■ 2
or
or
or
0-36 d=
0-72 0 =
72 9
72 ^ + 54 9-.
2X0-54 (100 — g)
4
= 0*54 (100—6)
= 54 (100—6)
54 X 100
or
6 =
54 X 100
5400
126
= 42*86®C
172 + 54 )
Fxamole 5 Equal bars of copper and aluminium are welded end to
^ If the free ends of copper and aluminium are rmmtaxned
%Tls::L 7e t:
0-92 and 0 5 calsfcmldegisec. respectively.
Let the temperature of the welded interface be 6. As the
bar is lagged therefore the amount of heat entering per secon
L nd requal to the amount of heat leaving the eopPer
interface. It is further equal to the he Q.
per second at the interface and ^ length of each
If A is the area of cross-section of the bar and x the 1 ngt
THERMAL CONDUCTIVITY 249
portion, then
For Copper bar ^ 0-92 ^ (lOO-O) ,.
For Aluminium bar
KA (e—e^) 0-50 A (e— 0)
^ X ^ X
0'92xA (100—0) 0*50 A (0-0)
^ ^ ‘ ~ ‘
X X
or 6 = 64. 8®C
Example 6. Calculate approximately the heat passing per hour
ihr^gh the walls and windotos of a room 5x5x5 metres if the, walls ore, of
hricks of thickne.ss SO ems and have windows of glass 3 mm thick and total
area 5 square metres. The temperature of the room is SO^C below that
of the outside and thermal conductivity of bricks and of glass is
I2xl0~* and 2oxl0~* e.g.s. units respectively.
Total area of the four walls including windows
= 5 X5 X 10^X4=100X 10^ cm^
Total area of all the windows
= 5 = 5 X 1 0* cm^
Net area of the brick walls=:ioo XIO^— 5X10^=95X10 cm^
Time =1 hour=3600 seconds.
Temperature difference between inside and outside»30®C.
Conducti^dty of bricks =I2X 10*^ e.g.s. units
Thickness of walls =30 cm.
Heat passing through the walls =
X
_ 12 X 10-< X95X lO^X 30X 3600
30
= 4104 X 10* cals.
Thickness of glass. =3 mm=o*3 cm.
Conductivity of glass = 25 XI 0 -* e.g.s. units.
Heat passing through the glass window
Q lO-^X 5X10* X 30X3600
^ 0^3
Total heat passing
= 410-4 X 10*4- 4500X 10*
=4910.4X10* Cals.
250 ENGINEERING PHYSICS-
11.7. Formation of ice on the surface of a pond. There is.
formation of ice on the surface of a pond because of the"
extraction of heat from water. When ‘-ice first begins to form,
on a pond, this is because of the removal of the latent heat from
a thin layer of surface water at 0®C by the comparatively cold air
layer above it. For further increase in the thickness of the ice
layer the latent heat is drawn from the unfrozen water by
conduction through the ice layer already formed. Hence heat passes
from water below the ice layer to the cold air above it.
Let X cm.be thickness of the layer of ice, t seconds after the
AIRAT-et
Fig. 11.3,
formation of the ice starts.
Let the thickness increase by an amraount dx after a further
time interval of dt second.
If a sq. cm. is the area of the surface, and p is the
in gm./c.c., the mass of the ice formed m the aje
dx=a dx P gm.
Heat lost by water =adx,pLc&\s.
where L is the latent heat of fusion of ice
This heat is conducted through the ice layer already present.
0
amount of heat conducted=.^o— cals.
where —is the temperature gradient through the ice and 0 C
X
6 .
is the temperature of the air above the pond surface.
From equation (») and («) we have
0
ap dx Ij=K A^dt
...(«•>
(i) Rate of increase of thickness of the ice layer as found fro
THERMAL CCNDUCTIVITY
equation (m) is
25U
dt pL X
and the time required for increase of thickness dx is
X dx.
If rj is the initial thickness of ice layer sec. after the commence-
ment of the formation and x^ is the thickness after sec., then bjr
integrating both sides, we have
k
X.
I*
dx
k k —
The time required to increase the thickness from x^ to is
pL
Ke
And time required to deposit a layer of thickness x from the
commencement of ice formation.
Ke
a;*.
Example 7. A pond is covered by a layer of ice 1 cm. thick, the
temperature of (he upper and lower surface of ice layer being respectively
—lO'^C and O^C. Find the time taken to deposit the next 1 m.w. layer-
of ice K^O'004 c.g.s. units, p of ice^O'928 gm.jcmK (Aero Society)
The time t take by ice layer to increase in thickness from x^ to a:*
IS given by
where p, Z, K and 6 have their usual meanings.
Substituting the various values, we have
*0-004 X 10 ^ ^
• =192 seconds=s3*2 minutes
*■ r increment of the thickness of ice
^1^?, wAen (fte tluckriess of ice is 20 cms. and the air temperature
MG. Thermal conduclmty of iee=0-00d cal/cmlsecrc, density
■25*
ENGINEERING PHYSICS
of ice —0’92 gm.jc.c. and its latent heat of ftision 80 calsjgm. After what
time the thickness will be doubled.
Rate of increase of thickness of the layer is given by
dx K 6
4
dt pL X
0
Where — is the temperature gradient through the ice and
— 6°C is the temperature of the air above the lake surface.
Temperature of air (atmosphere) 40°C
Temperature of water d^=0°C
0 = ^2— — {-40'^C) = 40°C ■ ‘
Thickness of ice x=20 cms.
dx_ K £
dt PL X
0.004X40
0 92 X 80 X20
= 108 X 10 ~^ cm./sec.
To find the time in which the layer of ice will be doubled i.e,, will
increase from 20 cm. to 40 cm. substituti the values of the quantities
given in the relation.
592x80 (40^-20:)^^^^
2 0-004X40X60 X60
Example 9. Calculate the time in which a layer of ice on the surface
of a pond which is already 3 cm thick will increase in thickness by 1 mm.
The temperature of air above the pond~ 20 C
The time t in which the layer of ice increases from a thickness
pL
we have
Xi to Xa is given by ^ 1 ^)
where p, L, K and Q have their usual meaning.
Substituting the values of various quantities given,
0-91X80 ^ (31“-3=) sec.
' ^ 0 005 X20'
= 222 seconds=3 minutes and 42 seconds.
11.8. Condnetion of heat through “any
the insulation on high temperature Several different
equipment are frequently constrac ? j conduction,
kinds of materials through which heat flows ms >. j
Take the case of a brick kiln wall. The inner layer consist
THERMAL CONDUCTIVITY 263-
. » • *
brick core of insulating brick and an outer layer of red brick. The
fire brick is used to protect the insulating brick from mechanial
abrasion and from the high temperatures existing within the kiln.
The red brick is cheaper than the insulating brick but has a higher
thermal conductivity and can not be used at the high temperatures
existing at the centre of the wall.
The rate of heat transfer by conduction through such composite
bodies can be calculated as follows. A compound slab consisiting of
two materials having different thicknesses and thermal conductivities
is shown in Fig 1 1.4.
The rate of flow of heat through section a is given by
Xi ...(t)
where 0^—6= temperature fall in the section a.
(b\ conditions the rate of heat transfer through section
1 ) IS also the same (because these are in series) and is given by
*1
X
2
or
or 0 ^—
^2
—3^ &e
or
X
[ ' +5- f]
•^>54
ENGINEERING PHYSICS
or
01^0,
El
^2
(
1 +
X:
K
a
X,
)
Substituting the value of d in either of the equations (i) or (li) we have
^ -^2
q=K^A
(
1
K.
Ki X,
)
1
q =
or
^2
r
k^a[.
Xa
K^A r
01-02
K.
«2
0 .
-e. h
a;g J
iC.
:>
I ^ 8^1 -^2
' X2
or
X
K
+
2
or
$i — 0z ^2
“ Xi X2 -2;a:
...(«)
Where g=the rate of heat transfer by conduction in X-cal/sec
or cal/sec.
K and ^,= the thermal conductivities of materials u and 6
^ evaluated at the average temperature of each m
_g'_-cal/m/°C/sec or cal/cm/°C/sec.
0 and 0-=the temperatures at the outside faces of the composite
^ body in ^C7. , ^
^=the cross-sectional area of the bodies taken normal
the direction of heat flow m sq. metre or sq.
, and *.=the thicknesses of materilas a and 6 in metres or cms.
This equation can be extended to include any
bodh. In «d.. by .dding .ddUi.nnl ^ t™ •• •>» °
equation (»h)*
THERMAL CONDUCTIVITY
255
11.9. Heat conduction through the walls of a thick cyliodrical pipe.
The section of a cylinderical pipe of length I whose thermal
Fig. 1 .5 END VIEW
conductivity is K and inner and outer radii are and respectively
is shown in Fig. 11.5. Let the pipe carry steam or some hot liquid
at temperature and let the temperature of its surroundings be
■^ 1 * Jliis thick pipe can be supposed to consist of a large number
of thin coaxial cylinders of increasing radii as shown in the end view.
It IS further assumed that heat is flowing radially across the walls
of the pipe.
Consider a thin cylindrical shell of thickness dr at a distance
r from the axis. Let the temperature drop across thickness dr be d0,
inen the rate of heat conduction through this shell per second is
given by
where .4=surface area of the cyclinderical shell.
The above relation can be written as
dr
q~ 2nl dS,
Integrating both sides between their respective limits, we have
. ’'2 ^8
q^^^~-27zKi^de
>*1
In the steady state q becomes constant since the quantitv of he'it
flowing per second across any thickness of the cylinder is the same.
eloge^ = _27t.K:2 (05,-5,)
or
log# r»
**1
256
ENGINEERING PHYSICS
or
iizKijdi—dz)
^■”2-3 logioTa
or
_ 2-3 log
10
2nl
wheee g=the rate of heat transfer by conduction from the
inner to the outer surface of the cylinderical thick
pipe in K — cal/sec or cal/sec.
it=the thermal conductivity of the material of the
thick pipe in jr-cal/m/®C/sec or cal/cm/^C/sec.
/—length of the thick cylinderical pipe,
and temperatures at the inside and outside of the
thick pipe.
ri and r 2 =tlie radii of inside and outside faces of the thick
pipe.
The examples of such cases are heat transfer through boiler
tubes or refrigerator pipes or through bare steam pipe^ec.
If the conduction of heat is steady then the rate of
fer through two cylinderical b^odier in series (thermal contact)
can be calculated by the equation.
r ]
2-3 ?
, -as as 'brrv' “S
iLterms to the denominator and the relation
tnc cy
additional
log
be statsd in the form
thermal conductivity
257
5 =
^2*3 logio
2tzI (5i— ^ 2)
rVr'
where 3=rate cf heat transfer by conduction from the inner
to the outer surface in iiT-cal/sec. or cal/sec.
and £^2= thermal conductivities of material (1) and (2)
evaluated at the average temperature of each in
^-cal/m/®C/sec or cal/cm/^C / sec.
f=length of the cylinders in metres or cms.
and 02=the temperatures at the inside and outside faces of
the composite body in ®C.
r/ and r/=the radii of the inside and outside faces of
material (l) in metres or in cms.
r 2 and r^'^the radii of the inside and the outside faces of
material (2) in metres or in cms.
/ conducted through a compound plate composed
different mater ials A and B of conductivities 0.32
ana 0-24 and each of thickness 3.6 and 4.2 cm. respectively. If the
temperature of the outer faces of slab A and that of slab B are found to
e s ^^dy at 96^0 and 8°C respectively, find the ^temperature of interface
A ana U and also the temperature gradients in A and B.
p,v temperature of interface A~B be as shown in
A B
Rate of heat flow through A —
_0-32^ (96— ..
jTg cal/sec . ... (i)
258
ENGINEERING PHYSICS
, , „ K^A ( 19 —^ 2 )
Rate of heat flow through B—
X 2
O-14^{0 — 8)
4*2
-cal/sec. ...(n)
where A is the area of the plate. Under steady conditions, the
rate of heat flow through the plates is the same. Therefore, equating
(i) and (if) we have
3*6
.*. Fall of temperature in
Thickness of .4=3.6 cms.
0 - 32 ^ { 96 - 5 )_ 0 - 14 ^ ( 0 - 8 )
4*2
0=72°C
^=96— 72=24°C,
Temperature gradiant in ^ 6*67 C/cm.
Similarly temperature gradient in
— ?=l5*4°C/cm.
4*2
111 The walls of a room consist of parallel layers in
Example U. The J thichness 2, 30 and 1 cm. rspectively.
wall per mmuU */ of thermal conductivities /or ceme»(
'Cuk::!Zd::Xodo7. o Je and O-OOOd C.y.s. units respect^ely.^
Rate of flow of heat through the compound waU
«=
X
KA
A (01-02)
'K^
cal/seC.
Heat flowing per minute
A (01—^2) — 60 Cals.
3
+ ^7+ K
3
10 * (30—5)
+
30
00*007'^ 0*006
1462*5 cals.
+
1
0*0004
, f mlnie of 2 '4 mm thickness
Example 12. A boiler ^thick. ^Surface area exposed to
coated inside with a layer t Calculate the maximum
■THERMAL CONDUCTlVlTr
• • •^•'1 I rill ritA
a59
of Copper and Ur, are 0-9 and 015 e.g.s. unite reepeclively. Latent heat
of steam at normal pressure is 5i0 caljgm. {A.il.I e!)
copper and tin
of Steam at normal -pressure is 540 caljgm. {A.3I I E )
Rate of flow of heat through the compound plate consisting of
transmitted per hour from the hot gases to water through the
compound plate, assuming the temperature of water to be lOO^C.
A [e^~~eA
- - X 3600 cals
100 {700—100)
~0^24 I 0*02
0*9 0'15
= 540x I0« cals/hour.
Heat required for evaporation of I gm of boiling water=540 cals
Maximum amount , of water evaporated per hour
^40X10®
540 8“S*
»^if^^7ltdone on'tZ tpoft^^Zr^'^ZZr '=”*
t^^s^e heat per sec. u,Uh the same. *ei»pera" are *
• •
or
or
•or
•or
*1
e,a +
EA
KA
E,A~~ EA
2’S , 2'S 2
0-1X25+0Tx55+0m1^ = ^
I 1
H
I
X25 KX6
+ — 1-
4 zKs
2K^K+iK=.i
3-5 ^=1
jr I
^ 37^ = 0-286
260
ENGINEERING PHYSICS
Example t4. Water U pumped through an iron ivhe 2 metres long
at the rate of 100 litresfsec. The inner and outer diametres of the tube
are 5 and 6 cm respectively. The outside of the tube is surrounded by
hot gases at a constant temperature of 500'^C. If the temperature of water
as it enters is 30°C, calculate the rise in temperature as it leaves the tube.
K for iron— O' 15 c.g.s. units, {A.M,I.E.)
This is an example of cylindrical pipe and the quantity of heat
flowing in this pipe, Q= —
2-303 logic ^
Thermal conductivity Z=0-15
Length l~2 metres:=200 cms.
Inner radius rj— 2*5 cm
outer radius rg — 3 cm.
Temperature of water on entering the tube=30°C7.
Temperature of water at out let =6^C
Average
temperature in the tube—
30-l-g
2
Temperature of hot gases di=500°C.
(
27tX0*15 X200 500
2*303 logic
2.5
30 + ^ \
2 '
cals/sec.
Heat taken
by water=mXi5xrise intemperature
=:IOOXIOOOXIX(0— 30) cals/sec
30+^
27tX0*15 X200
(soo—
)
= 10® (^ — 30)
2*303 logic-
2.5
Qj. 0 = 34‘9°G,
rise in temperature =(0—30)
= 34.9o_30=s4*9”C
Example 15 ^ater is pumped through an iron tube (£=p' 10)
metrfs lorj, at llte rate of 1000 inner
of the tube are 5 and 6 cms respeclwly. ‘ ^ eoO°C, the
mm thick {K=0 '004) is formed on the %ns%de surface of j
THERMAL CONDUCTIVITY
261
Quantity of heat flowing in the cylinderical pipe.
Q~ Kl{8^ — $ 2 )
2-303 logio —
where $i=600°C and $ 2 ==- — 0 being outlet
the water.
temperature of
27.X0-I6X2X100 ( 600- i2^)x60. cal/minute
2-303 Iog,0
2*5
This IS equal to the heat gained (taken) by water per minute
— m xs X rise in temperature
1000 X 1 000 X 1 X (6 -30) cal/mm.
= 10« (0—30) cals/min. ....
Equating {i) and {») we have.
^=66*3®C.
rise in temperature=(0— 3o)=66-3°— 30°
= 36.3®C
When a scale of
I
jQ mm is formed inside the pipe then it
becomes a case of two cylinderical lav^rc ^
for that the amount of Lat condicted is g%?n by^"
27,1 (g.— g,)
«=
2-303
Now
'■i'=V= 2*5 cm.
rj'=2*5— 0*01 cm.=2-49 cm.
>*a''=3 cm.
27CX 2X100 (eoo-!?^^
2*303 r — 5— W .
^^610 ... +
LOOO-4
Equating {%) and (i*).
2*49
0-1 6*°^“2^5]
X 60 cals/min.
• • •
(0
271X200 ^600—
^=30«76®C7
30 )
262
engineewng physics
rise in temperatures^^— 30=s30‘76--30=0*76®C
Example 16. A metal pipe having an external diameter 15 cm.
carries steam at 200°C. This is covered by a layer 2'5 cm. thick of
insulating material (K=0-0005). If the outer surface is at IOO°C,
calculate the heat loss per metre length per minute. Neglect temperature
drop in the metal. Also calculate the amount of steam condensed per
'^ninute. (^. M. /. B.)
Heat gained by the water flowing through the pipe
= 1000 X 1000 X 1 X (0—30) cal/min.
Now G=?^^^^^^l^^X60cals/miii
2-303 log,
Substituting Z=1 metre=:I00 cms.
r,=:^ = 7-5 cm.
2
r2=ri-|-thickness of layer n
^7*5+2‘5= 10 cm.
ex=20(fC
02=1OO°C
and 0*0005, we have
^ 27CX IOOXO'0005 (200— 100)
1 10
2-303 logic ^.3
= 6580 cals/min.
Q
Amount of steam condensed m—
_ Heat lost by steam pipe
Latent heat of steam
6580
540
= 12.2 gms/min.
Example 17. SUam at lOO'^G passes through “ 1””^
and outer diameters of 1cm. and 3 through the
immersed in ice. Find the temperature existing half uay mr^^
thickness of the wall of the tube.
Let e be the temperature half way through the thickness of wall.
and
ri=i=0*5 cm.
ra=l*5 cm.
^i=100°C
e^=o^c
and
THBRAL CONDUCTIVITY
263
^ lizlKie^—Oz) _ 2nlK{t00~0)
2-303 log,o ^ 2-303 logjo ^
Now considering only half the thickness of the tube, then
rj=:0-5 cm.
and
r2=0-5+i=0-5+0*5 = l cm.
6i=100®C
and
6= temperature half way through
^ 2 nlK{i 00 — e)
2-303 log„ i
Dividing (i) and (n),
1OO--0
log
xo
0-5
logic 2
0*3010
100—^
logio
1-5
0-5
logio 3 0-4771
= 0 631
100—^=63-1
or 6 = 100 — 63-l=36-9®C
11 . 10 . Experimental determination of Conductivity. The following
are the methods for determining the thermal conductivity for good
conductors and the poor conductors of heat.
(i) Searle’s Method. For good conductors like metal bars or rods
[ii) Lee’s Method. For good conductors like metal rods or bars
This method is quite suitable for the determination of the value of
K at different temperatures.
(ui) Cylinderical shell method. For poor conductors like glass or
rubber, ®
(iv) Lee’s disc method,
ebonite etc.
For poor conductors like rubber, glass.
(t) Searle’s method. The co-efficient
of thermal conductivity of good
conductors can readily be determin-
ed by means of Searle's apparatus,
shown in Fig. li.7, AB is a bar of
the material whose co-efficient of
conductivity is required. It is
covered with some bad conductor
e.^., felt or cotton to minimise
heat losses from its surface. The
end A is heated to steam
temperature by placing it in a
chamber through which steam is
passed. Alternatively this end may
be heated by an electric current.
At the end B is wound a coil having
Fig. ii.-i
264
ENGINEERING PHYSICS
two openings. Through this coil a steady flow of cold water is main-
tained. The temperature ^3 and 6 ^ of the water at the enterance and
exit is noted by the two thermometers Tq and placed in position as
shown. Two other thermometers and are placed a known distance
apart, in two cavities in the bar. To ensure good contact a little of
mercury is placed in these cavities. The flow of water is regulated in the
coil so that a steady state of temperature is reached. At this stage the
quantity of heat flowing through any section of the bar is the same.
The quantity of water flowing through the coil, (when steady
state has reached) in a given time, is collected and weighed. The
temperature of the thermometers Tx, T2, T^, and T4 are also noted.
Let m be the mass of this water which flows in the given f^ime t and
^3 and ^4 be the temperatures of water at the entrance and exit
respectively. Then, the heat which reaches the end B and is absorbed
by water in t seconds,
=m{ 9 ^—$ 2 ) cals.
Let d be the length of the rod between the thermometers Tx and
Tg ^d let 61 and Og be the temperatures recorded by them. If A is the
area of cross-section of the bar and K the co-efficient of thermal
conductivity, then the amount of heat passing through these points
in the given time t is,
- d
^3)=:
KAie^—Bxfl
required is taken in
a copper frame M as
n
(u) Lee’s method for good conductors. This method is generally
used to determine the conductivity of metals at various temperatures
The material whose conductivity is
the [form of a small rod. It is fitted in
shown in Fig. 11.8. A heating coil Cx is pro-
vided at one end of the specimen and two
platinum resistance thermometers Tx and T2
are placed near each of the ends at a known
distance d apart. To adjust the teniperature
of the apparatus a heating coil is wound
round the copper frame. The whole
apparatus is kept inside a Dewar flask.
The apparatus is immersed in
liquid air till the temperature falls below
the temperature at which ^ thermal
conductivity is to be determined,
liquid air is poured out of the flask and the
copper frame is heated by passing a current
through the coil C2 till the desired tempera-
ture is obtained.
The current is now passed tlmough tne
coil Cx and heating is continued till a steady
THERMAL CONDUCTIVITY
265
state is reached i.e., the thermometers and register a constant
temperature difference
If A is the area of cross-section of the rod and K its co-
efficient of thermal conductivity, then the quantity of heat passing
in t seconds, is given by
^ KA(e2-d,)t
d
where d is the distance between the thermometers Ti and T 2
EH
Heat produced electrically = — — calories in t seconds, where E is
the potential difference applied to the ends of the heating coil and
I is the current in amperes.
Under steady conditions the two amounts of heat are equal
Hence
Elt KAie^r-OyY
4-2
K
d
Eld
A'2A{d2-ex)
(tu) Cylinderlcal Shell Method. The apparatus for finding thermal
conductivity K of glass in the form of a tube is shown in the Fig.i 1.9.
Fig. 11.9
It consists of a thick walled glass tube TV t v v •
surrounded by steam by keepinl it in a slan^L ^ 'which is
chamber A. A steady stream oAvater is allowfd^^o^*^^
tube at a uniform rate. This is achieved bv olacinp a
oute^^Llber th? InleT? inThl
The water is heated
by
ihe heat given up by steam.
266
ENGINEERING PHYSICS
When Steady conditions are reached, the heat passing through the
walls of the tube is entirely taken up by water.
Two thermometers and are placed as shown, to record the
temperatures of incoming and outgoing water.
When steady state has been reached, then
T I T
Average temperature of water= ^ =$2
and temperature of steam=^i
If m is the mass of water flowing per second, then
Heat absorbed by water q=m {Tz—T^)
The inner and outer radii rj and of the glass tube are measur-
ed with the help of travelling microscope. Heat conducted through
the cylinderical pipe is given by
— ^2)
2-3
But q=:m (Tz—Ti)
m {T2 ^j) —
2nlK {$i — ^a)
2-3 log., ^
^1
Hence
K=
m (2^2 -T.) . 2-3 logi, ^
2 tcZ (^1—^2)
Determiaation of the thermal canductivity of a rubber tube.
The thermal conductivity of a rubber tube can be found by ^ first
passino^ steam through it for a short time. The tube is then picked
THERMAL CONDUCTIVITY
267
cooled several degrees below room temperature by addition of ice as
shown in Fig. 11.10(a). The temperature of water is then noted at
equal intervals of time as it rises from initial temperature to a
temperature N above the room temperature /j. From this graph the
rise in temperature per sec , (a'^C per sec. say) at the temperature
is obtained by drawing the corresponding tangent I to the curve.
When the water and calorimeter are at the same temperature as
the room {tj) no heat is lost by them to the surroundings. Thus the
quantity of heat flowing per second (Q/sec) through the rubber tubing
when immersed in the water is equal to the heat per second gained by
water and calorimeter, at the temperature t,
Qlsec=Wa cal/sec.
T7s=stotal water equivalent of the water
and calorimeter.
Qlsec=^KA, ^ -
Hence
Where
But
« •
Q/sec=iCTC (r^+ra) I
A=7t I is the area through which heat is conducted and
rj, and are the internal and external radii of the tube and I is the
length of XY.
dO
Thus
temperature gradient «
Wa==Kv:
~2 — ^1
0 — 0j^
The thermal conductivity K is calculated. ^
(iw) Lee’s method for bad conductors. From the heat equation
3— It is clear that in the case of bad conductors like
^ 9 would also be small.
A 3» either temperature difference and A should
be increased or thickness x should be decreased. As it is not possible
in ^*"'**' ^ increased by taking the material
m the form of a thin disc of large area.
tbin as shown in Fig. u.n consists of two
*> material about 0-1 metre in diameter and
con between four copper discs
the faces* of D and glycerine is applied to
copper discs to ensure good thermal contact with the
ENGINEERING PHYSICS
The copper discs Cg and C3 are fixed on the opposite faces of
heating coil The heating coil consists of a resistance wire
Fig. 11.11
wound on arnica strip as in an electric iron. It is heated by a
battery from outside. The temperature at each face of the disc
Dj is noted by the thermocouples di and and that at each face of
the disc D.^ by thermocouple ^3 and ^4. The whole apparatus is
suspended inside a constant temperature enclosure.
A steady state cuirent I amperes is passed through the
heating coil and the potential difference E across it is measured.
The heat so developed is conducted through and Cg and then
passes through and Dg. The copper discs Ci and are meant
for ensuring uniform distribution of heat so that heat flow through
Z>i and is normal and steady. When a steady state is reached
the temperature of the thermocouples B^, B^, B^ and ^4 are noted.
If the heat radiated from the edges of the discs C^, C^, and
£>2 is neglected, then the quantity of heat produced by the heating
• coil H per second is equal to the quantity of heat passine per second
through the discs and This heat is radiated to the atmosphere
by the copper discs and Cg.
Let the thickness of the disc be and that of the disc
' be x^-
Area of each disc —A
Thermal conductivity
Heat passing per second across
eals.
1 Xj_
f$ tf if if ff
* ^2
Total heat conducted away
KA (B2-B,)
. KA ( 0 s- B,)
^2
-w
Heat produced per second electrically by the heating coil
g=-^cals.
^ 4*2
•••(»■)
4-2
THERMAL CONDUCTIVITY
269*
Under steady conditions equations [i) and («’) are equal and hence
K can be calculated.
11.11 Convection. It is that 'process of transmission of heat in which
heat travels from one part of the body to another by the actual motion of
heated parlicals of the •medium.
The process of convection can be demostrated experimentally as
given below.
{%) Convection in liquids. Drop a few tiny crystals
of KMnOji near its side as shown. On
heating the water a stream of colour will be seen
rising from the bottom and then travelling in
the direction marked by the arrows. This can
be explained as follows.
Water at the bottom gets heated, expands,
becomes lighter and rises up. Cold water from
the top being heavier moves downwards and
occupies the position of water that has gone up.
Thus convection currents are set up.
Convection is possible only in liquids and
gases because convection currents cannot be
set up in solids. In fact, this is the only
important method of heating liquids and gases. The existance of
convection currents in gases can be shown by a simple experiment
given below.
(it) ConvecHcm in gases. Take a box"fitted with two card board
tubes P and Q. Place a candle with its flame under ther
tube g- If a smouldering piece
opening of the^tube P
the tube P and rises up in the tub^
brown paper is brought
smoke moves down in
as shown in Fig. 11. 13 . The
engineering physics
basis of this phenomenon IS the same as that of convection currents
in water The hot air in the tube Q being less dense rises up and
IS replaced by the cold air through the tube P. The movement of the
smoke shows the direction of the air travel.
Applications. Some applications of convection in nature and in
industry are given here under.
(1) l^atural applications of Convection. (o) Trade winds are
caused by convection. In hotter regions the earth’s surface gives
heat to layers of air in contact with it. These layers of air rise up
and are replaced by cold air from colder regions.
(fc) During the day time the cool sea breeze blows from the sea
to the land and at niglit the breeze blows from the land to the sea.
During the day the earth gets more quickly heated (because of low
specific heat) than water and cold see breeze replaces the hot air
•NVhich rises up from earth. During night as the earth gets quickly cooled,
the land breeze replaces the hot air from sea which rises upwards.
(c) Monsoons are also convection currents in air on large
scale. In summer, the Indian plains are very hot and the hot air
rises up. Moist air from Indian ocean flows tawards the land and
on getting cooled causes rain fall.
(2) Industrial applications of Convection {a) The ventilators
near the ceiling provide an out let for damp hot air exhaled by
the ^inmates of the room, while fresh air enters through the doors
and windows. In cinema halls etc. exhaust fans fitted near the
ceiling throw out the foul air while fresh air enters through, the
openings near the floor.
(6) The draught in the chimney of a furnance is also produced by
convection currents. A chimney above the furnance convects away
the hot air while fresh air enters the chimney from below the fire.
The higher and narrower the chimney, the better it is.
(c) Heating of buildings by hot water in cold countries.
{d) Radiators in automobiles. The heated water from the
cyclinders of the engine flows up through slanting pipes to the
radiator when it gets cooled by air currents and returns down to the
cylinder jacket at the bottom.
(e) Qas filled Electric lamps. A little argon gas introduced in
the glass bulb convects away the heat from the tungsten filament
uniformly in the bulb which enables the filament to get heated to a
higher temperature without any fear of melting and thus more light
is emitted by the lamp.
11.12 Radiation. We have already seen that conduction and convec/ion
are processes in which heat is carried from one place to another by tne
movement of the substance concerned. The heat from the sun which
reaches the earth, however passes through a considerable regi
which there is little or no material substance and hence heat can pass
through a vacuum. In this case we speak of the heat radiated by tne
ThCRMAL CONDUCTIVITY
271
sun. Radiation is the name given to the transfer of heat when the medium
takes no part in the transfer. The beat energy transmitted by this
process is called the radiant energy or heat radiation or thermal radiation.
Heat radiation resembles in many ways with light energy. The major
difference between them is their wavelength. Thermal radiation is of
lower wave length th^in light hence they are invisible to the eye. In
general the properties of radiant heat are similar to those of light.
Some of the properties are
(i) Heat radiation travels through vacuum with the velocity of
light (3x10® m/sec).
(m) Heat radiation like light travel in straight line.
(m) Heat radiation obeys the law of inverse squares like light.
(iv) Heat radiation is refracted according to the same laws
as light
(v) Heat radiation is refracted like light.
(m) Heat radiation exhibits^ the phenomena of interference and
diffraction.
(yti) Heat radiation can also be polarised like light.
When heat radiations fall on a body, some are absorbed,
some reflected and the rest are allowed to pass through
it. Those bodies which allow heat radiations to pass
through them are called diathermanotis. whereas those which absorb
heat radiations and are themselves heated are called athennanous. No
onaterial medium is perfectly diathermanous, air is only approximately
so. Rocksalt and Sylvine are diathermaneous to a large extent.
Water, wood and other liquids and solids are mostly athermar.eous.
11. 13. Black body. Regarding the affect of the nature of the surfaces,
it has been found experimentally that different surfaces all at the same
temperature emit radiant energy at different rates but there is a
rnaximum rate that none of them exceeds. It has also been found that
different surfaces absorb different fractions of the radiant energy
incident upon them. The hypothetical body whose surface would
.absorb all the radiation incident upon it is called a black body. Thus
,a perfectly black body is one which absorbs totally all the radiation of any
■.wavelength which fall on it. Since it neither reflects nor transmits any
•radiation it appears black whatever the colour of incident radiation
.ipay be. A surface coated with lamp-black
'Qr iplatinum black absorbs from 96% to
^8% of the incident radiations and may be
considered as perfectly black body for
practical purposes. On the other hand when
such a body is heated it emits radiations
of aU possible wave lengths. The radiation
einitted by a black body is known as Hack
body radtatxon or faU radiation or total
.radtat%on.
Fig. 11.14
272
ENGINEERING PHYSICS
Black body in practice. There is no such substance known which
exactly behaves like a black body as described above. In practice,
black body conditions can be closely realized by a small opening 0 in
a closed chamber say a hollow sphere as shown in Fig. 11.14
maintained at a uniform temperature. If the radiation enters through
the narrow opening 0 it is lost inside by successive reflections on the
inner wall and whole of it is absorbed. Hence the opening behaves
like a perfect absorber i.e., the opening or bole 0 (not the sphere)
constitutes a black body. The type of black body shown in the figure
was designed by Ferry. It is a closed hollow sphere painted black from
inside having small opening O and a projection P opposite to the
opening. This projection protects any direct reflection of the radiation
into the opening 0 from the surface opposite to the hole.
If the enclosure is heated to a uniform high temperature, then the
radiation coming out of the hole 0 are the full radiations or black
body radiations.
ll .14. Prevosf’s Theory of Exchanges. Before 1792 the ideas regarding
radiant heat were very much confused. It was thought that the hot
bodies were emitting hot radiations and the cold bodies were emitting
cold radiations. Prevost was the first to recognise that this was
incorrect.
He said that there was no such thing as cold radiation and proved
that heat radiation was essentially an exchange process- He said that
all bodies at all temperatures {above absolute zero) were emitting radiant
energy, the amount increased with the rise of temperature and was
unaffected by the presence of the surrounding bodits. The rise or faU of
temperature which is observed in a body is due to exchange ^ of radiant
energy with the surrounding bodies, his is known as Prevost's theory of
exchanges.
Near the fire we feel hot because our body receives
radiations than it gives out and near the block of ice, we
because our body gives out more radiations than it receives, e
considerations are quite general and may be applied to all sim
phenomenon.
It is quite clear from the above facts that when a t>ody
more radiant energy (radiations) than what it emits, the .
temperature and when it absorbs less energy, it falls in '
When the quantity absorbed is equal to the quan^tity
temperature of the body remains the same. When body has t
temperature as that of its surroundings it is a case of d^amc
equilibrium that is at this stage the body gives out ^ ^ the
of heat radiations as it receives from the surroundings.
temperature of surroundings rises or falls, Vi nnes It only
accordingly changes and the exchange of radiations continues. V
stops at absolute zero when all the molecular motion stops.
Hence it is concluded from the above explanation that the
absorbing power of a surface is equal to the emissive power-
THERMAL CONDUCTIVITY
273
11.15. Absorbing and reflecting powers. When heat radiation falls
upon a body, it is generally split into three parts : —
{i) a part is reflected in accordance with the laws of reflection.
(it) a part is irregularly reflected or diffused at the surface and
is thrown back in various directions and
{in) a part gets refracted into the body, of this, a portion may
get transmitled through the body, while the remaining may be
absorbed. The heat absorbed raises the temperature of the body.
The portion of heat absorbed or reflected depends upon the
nature of the surface. Those which absorb more heat than what they
reflect are called good absorbers while those surfaces which reflect
more heat than what they absorb are called good reflectors.
Absorbing or absorptive power. It is defined as the ratio of the
amount of heat radiations absorbed by a body in the given time to the total
amount of heat radiations incident on if in the same time. It is denoted
by aX.
If dQ is the amount of heat energy Ij'ing between wavelegths A
and A+dA incident per unit area of the surface, then the amount of
energy absorbed by it is equal to a\dQ. The rest of the energy
which is equal to (d(?—a;yc?Q) is either reflected or transmitted by
it. Hence the absorptive power a\ is defined as
Heat energy absorbed dQ^
Heat energy incident dQ
where dQ^—^x dQ
Reflecting power. If dQ-i — (dQ — °X(IQ) is the portion of the incident
radiations dQ which is reflected by a surface then the ratio
dQ
IS
called reflecting power of the surface. Hence reflecting power of a surface
is measured by the a7nount of heal radiations refiectvd by the surface to
the amount of heal radiations incident on it in the given time.
The complete study of the above facts state that the heat
radiation incident on the surface of a ^body ^from any source outside
the body is partly absorbed and converted into heat, partly reflected,
and may be partly transmitted through the body. The fractions
accounted for in each of these ways vary widely for different materials,
or example, both asbestos paper and polished copper transmit practically
one of the radiant energy incident upon them. But asbestos paper
a sorbs almost all of it and reflects verj' little, whereas polished
and reflects almost all. The fractions
® reflected, and transmitted vary also with the wavelength of
For example, glass transmits almost all the
shorter wave lengths but transmit very little
radation of the longer wave lengths.
274
ENGINEERING PHYSICS
From the definitions of the above items like abrorptive power and
reflecting power it is clear that those substances which reflect more
of incident radiations abrorb less. Thus we see that good absorbers
are bad reflectors. Dull black surfaces are bad reflectors and
good absorbers whereas polished white surfaces are good reflectors and
bad absorbers.
The absorbing power of a perfectly black body is unity as it
absorbs all the radiations incident on it. Its reflecting power is zero.
Emissivity of a surface is defined as the amount of heat radiated
-per second ■ y a unit area of the surface when the difference of temperature
between the body and its surroundings is 1°C.
The amount of heat radiated per second by a surface depends
upon the following factors.
(t) Area of the surface A
[ii) Difference of temperature between the surface and its
surroundings 6
{in) Nature of the surface.
♦ Q a ^ X0
or Q=A^ ^ X 6
where e is the emissivity*
The emissive power or radiating power of a surface is defined as
the ratio of the amount of radiations emitted by unit area of the surface
to the amount emitted second by unit area of a perfectly black body,
the emissions taking j)lace under identical conditions. The emissive
DOwer of a perfectly black body is regarded as unity. If we cqnfine
ourselves to the radiant energy emitted between the wavelength A
and A+(?A then in the limit dA-^O, we may write e^ for the
emissive power of the body for radiation of wave length A.
We must be careful not to confuse emissive power as defined above
with emissivity.
In the first instance the various properties of a substance relating
to radiation were defined in a relative manner. An exact definition of
emissive power is given below.
Thp emissive vower e\ of a body at a temperature for wavelength A
: JunedZ ZeU^gy radiked into vacuum per second per sg. cm. PJ
The absorpttvity a of a ‘gfven Tvavr^ength® ‘ that
the fraction of t)ie incident radiation of the g
the surface absorbs.
THERMAL CONDUCTIVITY
275
Grey surface- A gray surface is defined as one whose emissivity
is the same at all wave lengths and temperatures. Hence the ratio of
the rate at which a grey surface emits radiation of any wave length to
the rate at which a black body at the same temperature would emit
radiation of the given wave length is constant for all wave lengths.
Therefore for a grey surface.
The total rate at which grey surface emits radiation of
all wave lengths
^“The total rate at which a black body at the same tem-
perature would emit radiation of all wave lengths
None of the surfaces found in the engineering practice is actually
grey but the error introduced by assuming them to be such is usually
small.
11.16 Laws of Black Body^Radiations. The Laws of heat radiat-
ions are
(») Wien's Law.
(uj Kirchoff’s Law.
(t«) Stefan's Law.
(t) Wien's Law. It is a common phenomenon that the colour
of light emitted by a hot body changes as the temperature of the body
rises, first gloA^ing to a red then orange and finally white. The wave
length decrease progressively as we go through these stages of various
colours. Thus we see that the quantity of energy radiated is not
uniformly distributed over all wave lengths but is maximum at one
wave length say A„j depending on the absolute temperature T of the
black body. With the increase in temperature the value of wave
length which carries maximum energy is decreased. In other words,
\vave length corresponding to the maximum energy is inversely pro-
portional to the absolute temperature of the body,
Xm T— constant
[The value of the constant= 0*2898 cm deg].
This is known as Wien’s Law. It stales tluxt the wave length corres-
'ponding to the maximum energy is inversely proportional to the absolute
temperature. The curves in fig. 11.15, show how the radiant energy in
different wave lengths varies with the wave length A for different
temperatures of the black body radiator. At a given temperature
1650“^: for example, there is very little radiant energy Ex in the visible
spectrum but the energy increases to maximum as longer wave
lengths are reached and then decreases to zero. It will be seen from
the curves that for each temperature there is a wave length A„ for
which the ener^ in the spectrum is maximum and that the value of
Am decreases with the nse in temperature which is in. accordance with
276
ENGINEERING PHYSICS
Wien’s Law.
Fig. 11.15.
{ti) Kirchoff’s Law. From a consideration of the emissive and
absorptive powers of various substances it was observed that those
substances which have a high value for the absorptive power have a
high value for the emissive power. The relation pointed out between
absorptive power and emissive power is embodied in a law known as
Kirchoff 's Law. Thus Kirchoff’s Law states that at a given temperature
the ratio of the emissive power to the absorptive power for radiaiions of
a 'given wavelength is a constant for all bodies and is equal to the emissive
power of a perfectly black body.
To prove the above statement. Consider an enclosure whose walls
are maintained at a uniform temperature as shown in the Fig. 11. J6.
Let a body A of emissive power and absorptive power a^ be
placed in the enclosure, irrespective of the a initial temperature of the
body^, an exchange of heat will take place between b°dy
the surrounding enclosure and after some time a state of thermal
Fig. 11 16
equilibrium will be established between the two
THERMAL CONDUCTIVITY
277
Let dQ be the quantity of energy between the limiting wavelength
A and A+ d\ received in one second on a unit area of surface by
this body.
The body A will absorb a certain fraction of this energy that
is,an amount ax dQ, where ax is the absorptive power of A for a wave
length A. The remainder of this energy (1 — a;^) dQ, will be reflected
or transmitted.
Let ex be the emissive power of A for radiation of wavelength A,
and let Ex be the emissive power of a perfectly black body under the
same conditions.
The energy emitted in one second by unit surface of A by virtue
of its temperature is e^dA,
The total energy given out in one second by unit surface of
A = (l-ax) dQA'€xd>< ...(»)
This must be equal to the energy received dQ, when a state of
equilibrium is received.
{\^ax)dQ‘^exdX==dQ
In the case of a black body, the emissive power is Ex and the
absorptive power is unity
• *. Ex dx—dQ ...(iit)
Substituting this value of dQ in equation (u), we have
(l — a;^) Exd-XA-^xd^—Ex dX
ei^=ax Ex.
or
— ^constant for all substances
ax A
Which is Kirchoff’s Law. It is clear from the expression that if
€X is large then ax should also be correspondingly large which means
good radiators are good absorbers.
Applications of Kirchoff’s Law.
(a) If a piece of china clay which possesses some colouring material
in portions of its surface is heated in a furnace and is then taken to
a dark room, the coloured pattern will shine out more brightly than
the white background. This is because the coloured surface is a better
absorber than the white background, and in consequence also emits a
greater quantity of light.
called Z), and but when cold absorbs these very lines. This shows
^ ^ Sood emitter of certata wavliength is aUo
good absorber of the same wavelength. ®
5 the total radiation E from a black
body per second per unit area is directly proportional to the fourth
278
ENGINEERING PHYSICS
power of the absolute temperature T i.e.,
EccT^ or E=:<tT*
Where <7 is a constant known as Stefan's constant.
The value of o-=5*735 X 10® ergs /cm*! sec JK* (ergs cm'® sec'^ K~*)
in the C.G.S. system of units
and <7=5*735 X10-®joules/m®/sec/iC'*
^Joulesm'® sec-^ K-*) in M.K-S, system of units
= 5*735 X 10~® watts/m®/K* (watts m"® K-^)
because l joule/sec = l watt
As applied to a grey body having an emissivity of e, the above
law becomes
E=<TeT^
Suppose a black body at is surrounded by another black body
at T 2 . then
Rate of energy emitted by first black body per sq. cm. at
Rate of energy emitted by the second black body per sq. cm,
at
Hence, the net rate of loss or gain of energy radiation per unit
time per unit area
But the law can be extended to represent the net loss of heat by
a body after exchange with the surroundings and enunciated as
follows.
%
If a black body at absolute temperature is surrounded by another
black body at absolute temperature T^, the amount of heat lost by the former
per sec per sq. cm. is given by
E=\{Ti*-T2*)
Thus statement goes in support of Prevost’s theory of heat exchanges,
that the observed loss of heat is really the difference in the heat
radiated by the hot body and that absorbed by it from its surroundings.
Boltzman in 1884 gave theoretical proof of the Stefan's Law based
on tlie thermodynamical considerations. Hence the Stefan's Law is
also goes by the name Stefan Boltzman' s Law and <7 is called tne
Stefan Boltzman's constant. Boltzman showed that the law is strictly
applicable to all black body radiations.
11.17. Newton’s Law de<luced from Stefan’s Law. Considering a hot
body at absolute temperature placed in an enclosure at absolute
temperature T^, the rate of loss of heat according to Stefan s Law is
given by
- T,) (fi® + A*)
THERMAL CONDUCTIViTY
If the temperature difference {T^ — T^) is small i.c., Ti is nearly
equal to T^, the above expression can be approximately written as,
E=(T{Ti~Ti)X4T^^
=k{T^~T^)
if the enclosure is maintained at constant temperature 3 g, where
(a constant).
This shows that the rate of loss of leat is proportional to the
excess of temperature, provided the difference of temparature is
small, which is Newton’s Law of Cooling.
Example I. Calcxdate th'i maxirmLin amount of heat which may
be lost per sec. by radiation from a sphere of 10 cms diameter at a temperature
of 227^^0 when placed in an enclosure at a temperature of 27^C. Given
that a=5 7 wattslcm^fK*.
According to Stefan’s Law, energy radiated per sec. per sq. cm
is given.
E=a{T^^-~Tsf) ergs
Total surface area of the sphere = nr^=4 x 3*14 x 5*
Net rate of loss of energy=E x 4:rr- ergs/sec
a (T
.*. Heat lost/ sec= j = — -X 47tr* cals/sec.
where is the mechanical equivalent of heat 5 = 34*2 x 10 ^ ergs/cal
Now o-=5*7 X 10-'* watts/cmV/v^
or (r=5*7 X 10“® ergs/cm*/^'
Ti=227+.'173=500®A'
^"2=27 + 273 3 = 300 ® A.
.*. Rate of heat X >0 (500^-300 ^ )4 X 3*14 X 2 5
4*2 X 10^.
= 23.4 cal/sec.
Example 2. A black body with an initial temperature of 300°C is
alknoed to cool inside an evacuated enclosure surrounded by meltinq ice at
the r^e ofO'SS^ per sec. If the inass , specific heat and surface area of
are S'2 gm., 0-1 and 8 sq. cm. respectively, calculate Stefan's
According to Stefan’s Law the
given by ergs/sec/cm».
energy radiated/sec/unit area is
280
ENGINEERING PHYSICS
If A is the surface area of body, then total energy lost per sec.
IS given by, A(r(T^*~Ts^) ergs/sec.
— — —cals/sec. . . . (t)
Now heat lost/sec. from the body
£J=m5x rate of fall of temperature
= 32X 0-1 X 0-35= ri 2 cals/sec. ...(n)
Equating (i) and (if) we have
1-12 =
Now .4 = 8 sq.cm.
7\=273 + 300=573®A‘
^2 = 0 - 1 - 273 ® —21l°K
(T—
JX 1-12
4-2XI0’XM2
A{Ti* — T2*) 8 (573^—273^)
= 5-735 X lo'^ ergs/sec/cm^/^®^.
Example 3. A black body is placed in an evacuated enclosure whose
walls are blackened and kept at27°C. Compare the net amount of heat
gained or lost by the body (f) when its temperature is 227^C and (if) when
its temperature is — 7S°C.
According to Stefan’s Law of radiation, radiation energy emitted
or gained by a body is proportional to T^.
or E—cr where a is Stefan's constant.
Heat lost by the black body per sec. per sq. cm. of its surface
at 227®C or 500®A’ =(7(500)^.
Heat gained by the black body per sec per sq cm. from the
walls of enclosure at 27'C' or 300®Jf=a'(300)*.
Net heat energy lost/sec/sq cm of the black body is
/:„gl=:(r[(500)«— (300)«] =544X(l00)-*Xa'
= 544 X 10® X O'
(u) Heat emitted by black body when at ( — 73®^} or 200®^
=cr(200)^
Heat absorbed by black body from the walls is
= cr(300)'‘.
Net heat energy gained by ic from the enclosure walls
^ =(7 [(300)^- (200)«]=o*x 65X10®
Ratio of heat energy lost at 227®C to the heat energy gamed
544X 10®XO'
65X 10®XO'
at 73®C.
or
= 544 : 65
=41-85 : 5
Sample 4-. luminosity of Sigal star in orion ^mhllation IJfOO
times that of our sun. If the surface temperature of the sun ts 6000 K,
calculate the tempertture of the star*
281
THERMAL CONDUCTIVITY
Taking the sun and the star to be perfect black bodies, let T°K
be the temperature of the star.
Temperature of the sun =6000°^
Heat radiated by the sun per sec. per unit area=(r ( 6000 )*
Heat radiated by the star per second per unit area=o-r*
Since luminosity of the star is 17000 times that of the sun
o-r*= 1 7000 X cr (6000)*
1
r=6000 (17000)*
= 68520®K
Example 5. Calculate the surface temperature of Ike su7i and moo7i ;
^iventkat Xrn=d573 A.U. and 14 p respectively, Am being
length of the maxbnum intensity of emission.
According to Wien’s displacement law.
Am T— constant=0'2898
(i) For Sun.
Am = 4573 X 10-® cm.(l ^ .f7.= 1 0“® cm.)
Am r = 0-2898
or
T=
0-2898
4573 X 10“®
==6097*K
(n) For Moon-
Am = 14 A* * 1 4 micron
[l micron=10“® m.m. = l0~* cm.]
14 M = 14X 10“* cm.
0-2898
0-2898
14 X10-*
= 207®K.
Example 6. Two bodies P and Q are kept in evacuated vessels
niamfatned at a iemperatur^ The temperature of P is 327^0
to of Q ts 127 C. Compare the rate at which heat is lost froin P
For the body P
r = 327“C=600°.ff
To^irC = 300 °^:
£=o- (r*— 3V)
Ep=^<r (600*-300*)
We know
282
ENGINEERING PHYSICS
For the body Q
r=127'’C=400°/C
Tq^ll'^C = 300 ''£'
Eq^a [;T«-To*]=<r (400*-300^) .. (u>
Dividing (?) by (n), we have
Ep _<T (600* — 300^) _100* (6*— 3*)
E^ a (400*-300*) “lOO* (4*— 3*)
1296— XI 1215
= = = 6*95.
256—81 175
Example 7. Suppose that the maximum temperature in an atomic
bomb explosion is 10'^°K. What is the corresponding wave length of
maximutn energy 1
According to wien's law Ts^constant — 0-2898
, 0-2898 0*2898
— ij\ jq 7
= 2*9 X 10-* cm.
or
Am=2-9'*A.U.
Expected questions
1. (a) Define (i) Thermal conductivity and (•*) Thermal resistance. What
are their units in C.G.S. and M.K.S. systems. Find the dimensions of conductivity.
(61 Describe with experimental details Searle’s apparatus for determining
the conductivity of metals State the various sources of error and indicate other
methods by which they are taken care of.
2. Derive an expression for the rate of flow of heat through thick metallic
Describe Lee’s method for the determination of thermal conductivity of
(i) metals and rn) poor conductors. «
4. What do you understand by diffusivity ? How does it difier from thermal
5. Describe and explain the cylindrical shell method of determini g
condnctivity of a solid. Derive the formula used.
6. (o) How do you define the emissive and ^sorptive POwers of ^ body ?
lb) Name the three principal laws governmg the heat radiations irom
What is Stefan’s Law and Stefan's constant. Give the value of Stefan's
constant in C.G.S. and M.K.S. system of units.
Kwc^hX^fawV^d o' '-'3 I-'
CHAPTER XII
THERMOMETRY
12.1. Geoeral. The degree of hotnees of a body is known as its •
temperature. The science of measuring tamperature is known as,
Thermometry. In order to secure a reliable method of temperature
measurement, some property of matter is chosen which varies
continously with temperatiure. Solids, liquids and gases all expand
when heated, hence the amount of expansion or contraction of a
material can be taken as an index of temperature. Mostly the property
of expansion is utilised in the thermometers.
(<z) Thus a liquid enclosed in a glass tube of extremely fine
bore is a very practical form of instrument for measurement of
temperature. The alcohol and the mercury therrooinaters fall in this
category and are very widely used.
(it) A gas can also be used, but the apparatus required to
measure the expansion of a gas makes the method unsuitable for
industrial use, and its application is better limited to temperature
measurement in the laboratory. Constant Volume hydrogen thermometer
belonges to this type.
(6) The electrical resistance of metals increases when the
temperature of the metal increases. Hence the change of resistance of
a metal may be taken as a measure of its change of temperature.
I^he 2 >latinum resistance thermometers fall under this category.
If two wires of dissimilar metals be joined at their ends and
the temperature of one junction be raised above that of the other, an
elwtnc current will flow through the circuit, and the potential
mtterence between the two junctions may be used as a measure of the-
tamperature difference between them. This is a Seebeck effect, and.
this'^i thermometers or thermo-couples depend upon:
'noticed that all these instruments indicate the.
material constituting the thermometer, and there-
should be taken to ensure that the material has-
it §eing measured temperature.
ENGINEERING PHYSICS
Fixed points. At least two points of reference must be arbitrarily
selected or agreed upon, and the interval between these two points
can then be arbitrarily subdivided to give any desired scale of
temperature.
It is essential that the two points of reference should be such as
can be easily reproduced with great accuracy. For example, the
temperature at which a pure solid melts in the atmosphere is but
very little affected by small changes in atmospharic pressure. This
temperature can therefore be used as a point of reference. Any solid
can be chosen, an obvious choice is ice and the temperature at which
pure ice melts has been called the lower fixed point.
Lover fixed point is the melting point of ice. The second or
upper fixed point may be the temperature at which a liquid boils, and
here again water is the obvious liquid to choose.
Upper fixed point is the temperature at which pure water boils at
normal atmospheric pressure i.e. at 760 mm of Hg.
The difference between these two fixed points is called as
Fundamental Interval.
12.2. Scale of Temperature. For the purpose of practical measurements,
a temperature scale based on a number of fixed points that can be
reproduced with a high degree of accuracy has been adopted. There
are following types of temperature scales.
(i) Centigrade or Celsius Scale ('^C). This thermometric scale was
suggested by the Swedish Astronomer Celsius and is, therefore, called
the Celsius Scale, It is based on the freezing point of water taken as
zero degree and boiling point of water lOO^ both under standard
pressure. This scale is used in scientific work.
(n) Fahrenheit Scale TF). It is a thermometnc scale ^eviseid
by Fahrenheit on which freezing point of water is taken as 32 aria
boiling point of water as 212° both under standard pressure. It
used in engineering, medicine, and meteorology.
{in) Reaumur Scale rR). ^
freezing point of water corresponds to 0° and the boiling point of water
under normal pressure is taken as 80”. This scale is used for medical
and domestic purposes in several Europeon Countries.
{iv) Kelvin Scale TK) . The absolute scale (”.4) is also called as
Kelvin Scale. (°K). On this scale.
0°C=273®K or 273°*4.
T°A:=^f°C-f(273)
This is used in scientific calculations.
thermometry
285
(v) Rankine Scale i®Rn). It is the temperature scale which
corresponds to Kelvin scale but is based on the absolute zero of the
Fahrenheit System. Thuo
0° Rankine = — 460°i^
Relationship between Various Scales. The following equations give
the relation between various scales*
C _F~32_ R
100 180 80 *
In this way it is possible to convert a temperature reading from
one scale to another.
Also 273
Where C=temperature in °C
temperature in
temperature in ®Reaumur
A’saitemperature in ^Kelvin
R4=temperature in "Rankine.
12.3. Liquid Thermometers. Liquid thermometers are generally
known as liquid in glass tnermoineters and commonly used liquids
are mercury and alcohol. We will discuss only two types viz : —
(t) Mercury in glass thermometer,
(»*) Alcohol thermometers.
(») Mercury in glass Thermometer. It consists of a fine bore
glass capillary tube with a bulb blown at one end. The bore is
made as uniform as possible. The bulb and the stem of the
thermometer are filled with mercury by the method of alternate
heating and cooling. The bulb is then kept in boiling water lor some
time to expel traces of air and moisture. The open end is finallv
hermetically sealed. ^
By placing the thermometer in pure melting ice, the lower fixed
point can be marked on the tube. Similarly, by placing the thermo-
meter m steam, the upper fixed point can be marked. The interval
*7 lower fixed point and the upper fixed point is known as
fundamental interval and is divided into a number of equal parts each
<iegree which indicates temperature. The numLr of degrees
depend upon the scale chosen. agrees
•:86
ENGINEERING PHYSICS
On i\iQ Centigrade scale the interval is divided into lOO equal
parts the lower fixed point is given the value of 0* and the upper fixed
point 100°.
On Fahrenheit, scale, the interval is divided into 180 equal parts
the lower fixed point is given the value 32° and the upper fixed point
212 °.
Similarly on the Reaumur scale, the interval is divided into 80
• equal parts from 0° to 80°.
C£NTIGRAD£ fAHRSNHeiT R£AUMUR
Fig. 12.1.
The numbering may be extended above hundred and below zero.
The range of an ordinary mercury in glass thermometer is limited
tfrom — to 250 ^ 0 , as below— 30 °C crystals of mercury start
forming and above 250°C vapour of mercury collect m the u]^
space and suppress the level of mercury in the capillary tube. The
upper limit can be raised to about 600 °C by filling
mercury with compressed nitrogen, or CO, under be
this way the boiling point of mercury is raised. The rang
further inereased to about 750°C by using tubes of silica.
Mercury thermometers are generally employed 'ough work
as these suffer from a number of defects. If aS
use in very accurate work, then various corrections must be appne
to get the true tamperature.
THERMOTERT
287
The principal sources of error of mercury in glass thermometers are.
(i) Variation in the bore
(n) Effect of external pressure on the volume of the tube
{Hi) Effect of internal pressure
(it?) Change in the zero reading and in the fundamental interval.
(y) Exposure of stem of thermometer.
(vi) Composition of the glass.
Advantages of mercury as a tbermometric substance.
I. Mercury boils at 3ST^C and freezes at — 39°C. It remains
in the liquid state over a fairly wide range of temperature and thus
can be used to measure temperatures as high as 350*Candas low
-as — 3 2 ®C.
2. Being a good conductor of heat, it takes up heat quickly from
the body whose temperature is to be measured t.c. it is quick acting.
3. It has a fairly large coefi&cient of expansion i.e, the increase
in volume of a given mass of mercury for every rise of temperature
is fairly large.
4. Its expansion is uniform so that the increase in volume for
equal rise of temperature is the ssune.
5. It has a low thermal capacity, t.e., a very small quantity
•of heat is required to increase its temperature through a certain
range .
6. It has a very low vapour pressure at ordinary temperatures.
Thus the space above mercury contains very little quantity of
meraury vapour.
7. It is an opaque and a shining liquid and can be easily seen
an the glass tube.
8. It can be readily obtained in a pure state.
9. It does not wet the glass tube.
12.4. Alcohol thermometers. Mercury thermometer cannot be
rani U minimum
Jv. ° measure low temperature sometimes
Adv&ntsgcs* (t) Alcohol freezes
used to mels'^re
for ^e":? tlmpe° airi! SoJXrmo^te^^i^'thrfo^^
28S
ENGINEERING PHYSICS
more sensitive than a mercury thermometer to study temperature
variations.
(ni) The thermal capacity of alcohol is greater than that of
mercury. For a given volume, alcohol requires much less quantity of
heat than mercury to raise its temperature through the same range.
(iv) Alcohol is a light liquid and wets glass. Its motion is
therefore smooth even in a tube of narrow bore, while motion of
mercury is jerky.
Disadvantages. (0 Boiling point of alcohol is 78^C. Hence
alcohol thermometers cannot be used for the measurement ot hign
temperatures,
(if') Expansion of alcohol is not uniform. It increases with
rise of temperature. An alcohol thermometer, therefore, has to be
graduated with respect to a mercury thermometer
(lit) Alcohol is highly volatile so its vapours readily collect m
the space above and e.xert pressure.
(iu) Alcohol wets the glass surface. j.When the temperature
falls some of the liquid keeps on sticking to the walls.
12.5. Special thermometer. Besides the mercury in
meter some thermometer have been designed to meet requiremenu
special nature. These have been described in the following articles.
(i) Clinical thermometer [Doctor's thermometer). This ^
of maximum sensitive thermometer used for the temperature
of the human body. This type of thermometer .s gtaduaU''^
Fahrenheit scale having a range lyi^ between 95 F .
degree being divided into^ve parts. The normal averag P
of a healthy person is taken to be (98*4®F).
Fig. 12.2
It consists of a thin oHhe thermometer,
thick walled capillary tube 5 called ®
There is a constriction C in the bore near the bulb
When the bulb ot the thermometer « pla«d^
below thn tongue, the temperature of t b through the
expands. The force of expanding to indicate the
constriction C into the stem and thread
maximum temperature ot the human body. ^ falls and
When the thermometer retnoved constriction.
the mercury below C contracts, e J j c intact. When the
and leaves the thread ^bove the consmcmi^^^^^ held from the
thermometer 15 to be used mercury in the stem,
stem and is given a sharp ‘¥,y,b
passes tlirough the constriction into the bulb.
THERMOMETRY
289
(u) Six’s maximum and minimum thermometer. In this type of
thermometer the maximum and minimum thermometers are combined
into one piece of apparatus.
It consists of a long cylindrical bulb A filled with alcohol con-
nected to another small spherical bulb by a bent U-shaped
tube Ci> as sliown. Tlie bulb E also con-
tains alcohol which does not fill it com-
pletely, a space being left for expansion.
Part of the U tube CD contains mercury
which separates the alcohol in the tube
between A and C from the alcoJiol in the
bulb between E and D. The positions of
the mercury threads as at C and Z) are
indicated by small steel indices provided
with light springs to prevent slipping due
to their weight. These springs are not so
strong that they may oppose the movement
of the indices when these are pushed up
by the mercury thread. A magnified view
of the index is shown separately in the
diagram.
When the temperature rises the alco-
hol in the bulb A expands pushing the
mercury thread at G downwards and that
at D upwards. The mercury thread in
turn pushes the inde.x at D upwards
whereas the index at C remains in the
same position. If the temperature goes
on rising the index at D goes on moving Fig. 12.3
upward. Hence the scale on this side is marked with zero on the
slde^ highest temperature of I 20 ®if’ on the upper
When the temperature falls the alcohol in A contracts. The
mercury thread at C rises upward and pushes the index. The
mercury thread inD falls downwards and the index at Z) remains
innlvV position. If the temperature goes on decreasing the
index C goes on moving upward. Hence the scale on this side is
marked with zero at the top and maximum of iiO^F on the lower side.
indicl?rd by the"*^Xn ^on 't& D and‘th” ^
temperature by the positiou of the ^ndex on the sMe C
«ith?h: mlrcurlurlace doL
Iwrse-shoe mag^t ^ strong sma
290
ENGINEERING PHYSICS
B
{in) Beckmann's Thermometer. Beckmann's thermometer is a
special type of thermometer used for measuring small differences in
temperature, correct up to \o°0 or cases
where the change in temperature is important,
and the knowledge of the actual temperature is
not necessary. It consists of a large bulb A
containing mercury and connected to the capi-
llary stem B, graduated with 5 or 6 large
divisions each corresponding to 1®C, each
division subdivided in to 10 or 100 equal parts.
To use the thermometer over a large range of
temperature, a constricted bulb C is provided at
the top end of the stem. The bulb C forms a
chamber into which some of the mercury may
be transfered or vice versa, in order that
sufficient mercury is left in the bulb A to
enable the lower temperature to be read, the
level being near the foot of the scale at the
start. The bulb A is immersed in the vessel
whose temperature change is required, the
higher temperature is then read. The difference
in the readings gives the required difference
in the temperature.
The advantage of this arrangement con-
sists in greater accuracy being obtained whith-
out the necessity of employing a very long
stem, which gets broken easily.
Other suitable liquids for thermorneters
are acetone, aniline, glycol, methyl alcohol.
firl Standard constant volume Hydrogen
Thermometer. This thermometer '
of the changes of pressure of hydrogen while its
volume is kept canstant.
pressure. The manorneter consis connected to the reservoir R
by'rrs"orarJbberW contain^
with a barometer C. 'hHem orwHch d^ps into
V
Fig. 12.4.
THERMOMETRY
291
A scale is placed vertically so that the zero of the scale coincides with
ii-ig. rz.5.
the platinum tip Pj. The position of the platinum tin p ic u
the vernier V. which can slide along the scale. ^
•u perform the experiment surround the • i
hulb C with ice. The position of the reservoir P is adiustin ^
the mercury in the tube A touches the platinunl tip P
barometer D is adjusted by moviner it ud and dnwn i* Then the
so that the mercury in it touches the platinum tip P Th^ r® scales,
the vernier F is noted along the scale. This ffives th^ reading of
gas in the cylinderical blub C at 0°C If ho is thp ®
between top of mercury in ^ and B then L-P^h difference
atmospheric pressure, ® +«o* where P is the
(tt) The cylindrical bulb O is now surroundpH ^
expands pushing down the mercury in the^tuL hydrogen
the reservoir B and that of the barometer C arl 'f; position of
S', Ksw i£”'s.p?:“
• * *4
292
ENGINEERING PHYSICS
Hence ■Pxoo=-^+^oo where AioQ=vertical difference between
A and B.
{Hi) The bulb is now surrounded by the bath whose temperature
is to be found out and the pressure Pt of the gas in the cylinder G at
the unknown temperature is again noted after making the necessary
adjustments,
Pi=P-\-ht where ^i=vertical difference between mercury leves
between A and B.
Calculations. If y is the coefficient of increase of pressure at
constant volume, then
Fi«,=P„ll+yl00)
■^100 Po
or
' FoXlUO
and
Pt-Pf>
or
PoXt
From (i) and (it)
^^ioo--Po Pt-P>
PXIQO PoXi
X 100.
•* 100
{P+ht)-{P+ho)
X 100=
hf — ho
X 100
Corrections. For accurate measurement of temperature, corrections
must be applied for the various sources of errors.
(f) The volume of the bulb C is eoXn^
tures so that the measurements are not made stnc y
volume. .
(ii) Some gas is absorbed iu the walls of the bulb especially at
low temperatures. +hp density of mercury
liii) The barometer pressure changes as the de y
changes with the change in temperature.
,iv) The hydrogen gas is not perfect and deviates slightly from
the behaviour of a perfect gas. oq ^nd
Advantages. (0 has a wide range from-220 to
with Nitrogen it can be used upto 1600 considerable, the
(2) Since the expansion of hydrogen 6^ ,
thermometer is very sensitive. It has an accura y
(i) o-OOS»C between fixed points f.e. between 0(7 to too .
THERMOMETHY
293
(n) 01®C upto 500®(7.
(iti) upto 1600“C.
(3) Unlike mercury thermometer, readings on different gas
thermometers using the same gas agree among tlicmsclves. This is
because,
(t) gas can be obtained in pure form.
(t?) expansion of the cylinderical bulb does not materially affect
the readings.
(4) Its readings agree with thermodynamic scale (of Lord Kelvin)
which is the ultimate standard of reference. That is why tliis
thermometer is quite suitable as a standard of reference.
Disadvantages.
(i) The thermometer is slow in action*
(it) It cannot be used for the direct measurement of temperature
as its bulb is too large.
(iii) It cannot be used for measuring varying^temperatures.
Example The pressurp.^ of lh<i gas in a con<!tant volume qa^
thermometer are 100 cms. ar,d 230 09 cm^. of mercury at O^C and 100^0
^ pressure is 125.8 cms.
/ mercury. Calculate the lempira*ure of the ba'k.
Po = 100cms.
^ 100=136*99 cms.
125*8 cms.
If t^C is the temperature of the bath, then
Pt-Po
Pioo-P.,
125*8 — 100
736 * 99—100
25*8
X 100
X 100
3^;^ X100=69*74®C.
the instruments used are called *^"'P“*'ature, and
designed and developed for temperatu^i^?®>f”' Principally
measured by the usual mercurv beyond those that can be
used, to measure moderate as weU as tempir"atures?'
Pyrometers-in general usp
to .300«C.
294
ENGINEERING PHYSICS
(ii) Thermo electric pyrometers for temperatures up to 3000®O.
The example is thermo electric thermometer.
{Hi) Pyrometers which measure high temperatures upto the order
of 3 200®C by the method of radiation. There are two types in
practice :
((i) Total radiation pyrometers, range is up to 1500®(7.
(6) Optical Pyrometers, range is up to 3200''C'.
(j) Platinum-resistance Thermometer.The first resistance thermometer
was constructed by Sir William Siemens in 187]. The platinum
thermoneter is due to Callender and Griffths and it gives quite reliable
results over a wide range of temperatures. Platinum has been closen
as thermoetric substance because of,
(i) its high melting point
(u) Chemical stability at high temperatures
{Hi) its availability in pure state, and
(tv) its large temperature range.
Principle. The variation of eletrical resistance with temperature
has been made use of for the measurement of temperature m a
platinum resistance thermometer. The resistance of pure platinum
is found to vary according to the formula
where a and p are constants and Rt and Ro are the resistances at
t°C and 0°C respectively.
For pure platinum a=3.94 X 10~® and
p=— 5*8X 10"^
The values of a and ? are determined by finding the resistance of
the platinum coil at.
{a) Melting point of ice i.e., at 0°C
{b) Boiling point of water i.e., at lOO^C and
(c) Boiling point of sulphur ue, at 4 44.6“0.
Clausius showed that for a conductor
i?(=:i2o (I ^
where Rt is the resistance of the conductor at i C
Ro is the resistance of the conductor at 0 C
eood Now if the resistance i?o, i^ioo ana oi a pie t'
fre found at lOO^’C and /“^.respectively, then
Rt=Ro
or — Ro—^o<^^
(l+« 100)
THERMOMETRY
295
or i?ioo— i?o=i?oa 100
Dividing (t) by (i7) we have
-^100 _ ^
■Ro 1 00
...(u)
R
100
or
Rt~~Ro
t =^ — xioo
This value of t is correct on what is called Platinum scale and
this temperature is termed as Platinum temperature denoted by ip.
Rt — Ro .....
X 100
The quantity R^^ — Ro is known as the fundamental interval of
the thermometer.
The temperature on the phxtinum scale differs from the temoe-
rature on the perfect gas scale or absolute scale. ^
r- ,1 temperature t from the observed temperature /
Callender has deduced the following relation peraiure tp.
^ lAioo/ looj
• • •
{iv)
wire^er DsV.Z‘u‘ particular specimen of platinum
wch^vr* reUVion (iv)^
Rt-Ro
■^100 — Ro
-xloo f^^. Rt—Ro
X 100
or
and
Rf
Rt — Ro
R\wr^Ro'
Ro (ocf+pf^)
^(100 a-t-{100j»
Ro (l+of+p/*)
■Ro (a« + pi®)
Ro (a lOO+P (100)2
(g+pf )
(a+IOO 3)
“«-f 100 3 {«+100 3)-/ (a + p^)]
+7o^ (*0® <-<*)
r/_LV f t x-|
«+i00 3 Lboo )
®[(ioo) (f^)]
where
» 3 (100)2
a + lOoT
296
ENGINEERING PHYSICS
As p is negative quantity 5 is positive.
Construction. It consists of a platinum wire free from impurities like
silicon, carbon, copper, tin etc, doubled on itself to avoid induction
T effects, wound on a thin plate of mica, enclosed in a
porcel.iin tube as shown in Fig 12.6. The ends of the
platinum wire are attached to platinum leads which pass
through holes in mica discs closely fitting in the upper
part of the tube. The free ends and Tg leads
are connected to the terminals at the top of the instrument
for connections to the external circuit, Th mica discs
are used for insulating the wires passing through them
and for preventing convection currents from moving up
and down the tube.
As the lead wires are long enough, they offer
appreicable resistance whicli will increase with temperature
and so introduce errors. To compensate for the rsistance
of leads, an exactly similar pair of leads Cj, with their
lower ends joined together are placed close to the
platinum thermometer leads as shown. The tube is
sealed at tlie top to protect the platinum coil from the
effect of moisture etc.
Working. The resiestance of the wound platinum
coil is found with the help of a modified form oi Wheat-
ston's Bridge as shown in Fig. 12 . 7. The ratio arms P and Q
are of equal resistances. The leads Tj, are connected in
the arm X and compensating leads C^, Cg are connecte
in the arm The resistance of platinnm
changes roughly by 0.01 ohm per degree
cantigrade. In a resistance box, steps
of such a low resistance are not possible,
hence a wire of uniform area of cross-
section, of length 2a and resistance P
ohms per centimeter is inserted as
shown. Such an arrangement is known
as Callendar oetw? Griffth s Bridge.
When platinum wire is at 0'’C, value
of R is so adjusted that the balance
point is obtained at the mid-point of
wire la. If the values of resistances
in arms P and Q are kept equal, then
Resistance in X arm=resistance in
Fig. 12.6
R arm.
X+resistance
compensating leads
or
Ro=R—X
of
THERMOMETRY
297
When the platinum coil is placed in a bath at its resistance
increases from J?o to The balance point will then shift towards the
side of the platinum thermometer thereby decr.'asing the resistance
on the side of X and increasing the resistance on the side of R by an
equal amount. If a new balance point is obtanied at a distance x
from the middle point which is marked zero, then
Rt + {a—x)p = R-^{a-\-x)p.
Knowing x and p, Ri can be calculated. Similarly the resi'^tance of
platinum thermometer at t OO^C ^lon is found by placing the
thermometer in a steam bath.
Knowing Rq, R, and the temperature on the platinum scale
is found by using relation. (Hi) and tlicn true temperature is
determined by using relation {It') if the value of 8 is known.
Range of platinum Thermometer. This thermometer can be
reliably used from — 200 ^C to 1300®C. Its readings arc accurate to
0.0 rc upto 500*C and to O l^C upto i 300^C.
Advantages.
1. It has a wide range t.c., from— 200*^0 to 1300'^C.
2- It is correct to 0 0l°C up to SOO^’C and to 0*I°C7 up to 1 300’C.
3. It is very convenient to use and when once standardised by
comparison with a gas thermometer, it serves as a reliable secondary
standard of reference.
4. It is free from zero error, because the resistance of the pure
and well annealed platinum wire is alw«ays the same at the same
temperature.
5. ^'Stem Correction" has been eliminated in it by the use of
compensating leads.
6. The temperature can be measured sitting quite far off from
the furnace etc, by just increasing the length of the leads.
Disadvantages.
1. It has large thermal capacity.
2 . Porcelain being not a good conductor, takes a long time to
attain the temperature of the bath and is therefore unfit to measure
rapidly changing temperatures.
3. It IS not suitable for measuring unsteady and raoidlv
temperatures. '' t' y
varying
‘“*1® I® suitable for
measuring at a point on a hot source.
5. The tube is fragile and cannot with Stand mechanical shocks.
Run in a platinum resistance thermo-
meter are ^ 585, 3 510 ohms respeclitiely. When it was vlaced in a furnace.
298
ENGINEERING PHYSICS
the resistance R was found to be 9-098 ohms. Calculate correct ud to PC the
spdZeZf XutriTl-T S /<«■
Ihe temperature on the platinum resistance scale is given by
Now
4 Rt~~‘Ro
w X 100
■“100 — “O
i?o = 2*585fi,
-ffloo=3•5l05^
i?/=9'098^t
, _9-098-- 2*585
^ 3*510 — 2^5^^*®®
= 704°C.
Again
704 = 0*00015 i2_o*0]5 t
0*00015 «2_i-015 <+704 = 0
or
or
way.
This is a quadratic equation which can be solved in a usual
^_1015±^(i.0l5)*_a X0 000I5 X704
2X0-00015
1-01 5 + 0*78
0*0003 0*0003
«5984®C' or 784‘’C'.
Example. 3. A fixed mews of a gas maintained at a constant 'pressure
occupies 200.0 c.c. at the temperature of melting ice. 273.2 c.c. at the
temperature of water boiling under normal pressure, and 525-1 c.c. at the
normal boiling point of sulphur. A platinum wire has resistance of
2 000, 2-178 and 5-280 ohms respectively at the same temperature.
Calculate the value of boiling point of sulphur given by the two sets
of data.
(i) On Constant pressure thermometer
Vo=200 c.c. Fioo=273*2 C.C., ^^=525*1 C.C. where < is the
point of sulphur on the gass scale.
Vt-Vo
boiling
<=
X100 =
525.1—200
Vieo—Vo 273*2—100
[ii) On platinum resistance thermometer,
Ro~2.00 SI, i?ioo=2*778 St and i?,= 5*28 5^.
X100=444*2X.
thermometry
... <=The boiling point of sulphur on the platinum scale
299
Ri — Rq
•^100"
X 100=
S-280-2-000_ xl00=421-6'’C
2*778 — 2*000
12.7. Seebeck efifect. Tjus^enmnen^ was
12.7. seenecK eneci.
in 1821. litw^ssimUiar me^^^
in 15 ^ 1 . Ai twu ui3»uiiixio.4 v ?. /^r^\A
jtui^ns aTiK^ irFig.iz.S and on^onKfTunctiq^^
while the other is heated, an electric c^ entg.taits. flowing througii
the metals as indicated by“ a glHvanor^t^ar^hence an e. m
developed across the two junct^ons^*^-'—" ^ J »
Cu^ —
Fig. 12.8.
f current flows from^u^to Fe at the hot lunct ion B.
is called thernw^eleclrtc^urreni and the corresponding E.M.F.
The
current ^
is known as Thermo-e.m.f. The arrangement is k nown as thermo-
coople. TJie production of thermo e. wi. /. by o t tiermo couple when o ne
jwnctxons ia heated keeping other cold kn own a9
rne thermo e. m./i ^nerated depends upon
the
eci
(t) Nature of the two metals forming thermo couple
(u) Temperature difference between hot and cold junctions.
r (.
It is prefarable to measure e. »»./., instead of circuit current:
because the c. m. f. developed depends on the temperature difference;
only where as the current depnds upon the resistance of the circuit
which alters considerably with temperature.
Effect of temperature. If the graph is plotted between thermo*
c. m. /, E and temperature t,it is of the parabola type as shown in Fig.
12,9 and it obeys the relation.
E=at-\-bt^
where a and 6 are constants of the metals forming the thermo couple.
i<! Clinic temperature of the junction B'
For cmili the thermo-clectric current continuously increases..
to t.mr>i>r t «•»» /• is proportional'.
increlTe^ and fo; difference of temperature e. m /. first;
increases and for a certain temperature of the hot junction it becomes.
300
temperature is known as neutral temperature and is
It TconsZfr eurrem
mofo’io I ^ constant for a given couple and depends upon the nair of
metals chosen but is independent of the temperature of the^Told
co'upl‘e°"s' temperature for Cu-Fe thermo
couple IS 270 C whatever the temperature of the cold junction may be.
If the temperature of the hot junction is further increased the
thermo electric current (or thermo H.m.f.) decreases and finally
becomes zero at a temperature known as tlie t'^mporature of inversion.
The temperature of inversion is as much above the neutral tempe-
rature as the latter is above that of the cold junction. For example
when the temperature of the cold junction is O^C the temperature of
inversion is 540®C but if the temperature of the cold junction is 40°C
the temperature of inversion is 500®C for Cu— Fe thermo couple. If
the tempjrature is further increised the thermo electric current
increases in the opposite direction as shown.
The thermo couple if worked on the rising portion of the curve,
where the thermo e.m.f. is almost linear function of the temperature,
it can be easily adopted as a thermometer to measure any unknown
temperature.
(i) Thermo-electric Thermometer. Although it is inferior to platinum
resistance thermometer for temperature below but is much
more sensitive and reliable for higher temperatures. The maximum
temperature up to which it can reliably measure is 3000°(7 which is
much higher than platinum thermometer.
Principle- It makes use of Seebeck effect or thermo electricity
which has been explained in the preceeding Art 12.7 ^
thermo couple as a thermometer following points have to be taKe*^
into consideration.
THERMOMETRY
301 .
1. Selection of suitable metals. Choice of metals for the thermo
couples chiefly depends on the temperature to be measured. The
metals chosen should be such that the range of temperature for which
they are to be used should be quite lower than the neutral temperature.
Some of the couples in common use are
(a) Base-metal Couples.
(i) Copper-constaniaii and iron-constanlan couples do quite well
from 200®C to 400®C, both in the laboratory and for industrial work
because they develope large t.m.j. of the order of 40 to 60 and
their tempeiature versus e.m.f, curves are practically linear.
(n) Iron-nickle couples can be used up to 800'^C. Tlieir tempe-
rature versus e.m.f. relationship is also nearly linear over this range.
(m) Nickle chromium, S'icklp-aluminium couple. Nickle chromium
consists of 90% of Ni and 10% Cr ; Nickle aluminium consists of
about 95% Ni and 5% A1 with Si and Mn. This combination has the
commercial name of chromel — Alomel thermo couples and can be used
up to i200®O.
[Note. Small quantities of Si and M n are necessary to reduce the
brittleness at lower temperatures.]
The advantages of using these metals is that they give a relatively
large e.m.f. per degree and their co^t is low so that heavy sections can
be empl.jyed, giving low resistance.
For high temperatures however, I200^C to 3000®C these base
metals are not satisfactory as they are easily oxidised and Noble-
Metal couples are used.
(6) Noble-Metal Couples.
(fv) Platinum-irridium or Platinum-Rhodium couples may be used
upto 1750®C.
^ Thermo-Couples of irridium and an alloy of irridium and
rubidium (90% Ir and io% Ru) can be used upto 2100‘’C.
(v») Tungsten — Molybdenum couples are the best for higher
temperatures upto 3000'’a.
anH carefully protected from reducing gases
nar are rather costly and the e.m.f. developed
P degree is about fifth of that couples made of base metals.
^ thermoelectric thermometer two wires of
e materials depending upon the temperature to be measured are
-302
ENGINEERING PHYSCISCS
welded togettier at one end. The welded end forms
the hot junction. The two wires are insulated
-from one another by passing them through thin tubes
of clay or hard glass and are threaded through
mica discs to keep them in position. This is
further enclosed in an outer protect! ig tube of
porcelain, quartz or hard glass depending upon the
temperature to be measured and provided with a
box wood head fitted with terminals and to
which the wires forming the thermo couple are
connected.
To these terminals are connected flexible com*
pensating leades of the same materials as the
elements of the couple itself. In this way the
cold junction is transferred to a convenient distant
place where a constant temperature generally
is maintained as shown in Fig. 12.10.
Working (t) For obtaining higher accuracy the
thermo e.m./. is measured with the help of a potentio-
meter and a standard cell. The connections are made
as shown in Fig. 12.11 where ^5 is a potentiometer
wire of resistance 1 ohm per metre. With the plug
placed in between the studs h and c of the two way
key, the standard cell is first put in the circuit and the
balance point for it is obtaind, say at D, on the
potentiometer wire by adjusting the values of the
resistance and so that, total resistance i?i-f
that of the portion AD of the potentiometer
wire
= 1018.3 ohms.
CLAV
TUBES
'MICA
0I5CS
PORELAiN
OR
QUART 2
TUBE
OR
HARO
glass
HOTENO
Fig. 12.10
Then, since the e. m. f. of the standard cell is I *0183 volts, it is
•clear that the current flowing through the potentiometer wire is
1’083 ,
= 1 milli ampere
1018*3 ^
Therefore, obviously fall of potential across one length of the wire
= 1 raill-amperex l ohm— lo-^xi Volt
= I mitli— Volt.
And, therefore fail of potential across l mm. length of the
potentiometer wire
= — — =10“* volt = l micro-Volt.
1000
It is clear that potential differences of the order of I micro-volt
•can be measured in this manner.
THERMOMETRY
The standard cell is now thrown out of the circuit and the
thermo-couple, with its hot junction in the bath or in contact with
STANDARD CfU
Fig. 12.11.
the given body and its cold junction at 0®C, is taken into circuit by
putting in the plug between a and c instead of between b and c, and
the balance point on the potentiometer wire obtained as before. Let
it be at D‘, at a distance I min from the zero point A.
EM.F of the thermo couple— ^ X10“® Volt.
Thus with a potentiometer the can be
up to a micro volt which corresponds |
to about 10®(7 for Platinum, Platinum
measured correct
Rhodium thermo couple and —
for a Copper constantan thermo
couple. To deduce the temperature
from the experimental value of c. m. /.
a calibration curve is generally
supplied by the manufacturers. The
caUbration curve is drawn between
thermo e. m. /. and temperature of
the hot junction. The caliberation
curve of the thermo couple is drawn
as follows. The hot junction of the
thermometer is put in a bath whose
a-
r£MPeRATUR€
Fig. 12.12.
temperature can be varied. The e.w. /. so developed by the thermo
couple is measured as discussed above. The caliberation curve as.
sho\yn in Fig. 12.12 is then drawn. After this has been done for a
particular thermo couple, then an known temperature can be read
off corres ponding to the e. m. /. at that temperature.
ENGINEERING PHYSICS
(u*) To determine the temperature of the hot junction the emf
developed by the thermo couple is measured by milli voltmeter
graduated to read the temperatures directly. The milli voltmeter
used should be of high resistance so that the increase of resist ance
of the leads or thermo couples does not effect the total circuit resis-
tance appreciably. Though a direct reading and convenient arrange-
ment. Its accuracy is limited to within ± 5 '’C'. It is therefore,
suitable only for industrial concerns where no great accuracy is
required. For higher accuracy, it is necessary to use a suitable
potentiometer as discussed above in detailes.
Advantages. (») They are quite cheap and can be easily cons-
tructed. Hence they are frequently used for laboratory work.
{ii) Their thermal capacity being small, there is no time lag and
they are very well suited for measuring rapidly varying
temperatures.
[Hi] They can be used for measuring temperature at a point.
(iy) They can measure temperatures over a wide range from
— 200 ''C' to 1600®C and even extending upto 3000°C with suitable
thermo-elements.
(y) They are best suited for remote indication and recording.
Thus one junction of the thermo couple may be at the top of the
mountain and the other at a convenient place in the valley below.
(y*) Tliey can be made direct reading.
Disadvantages, {i) Each thermometer requires a separate calibe-
ration as there is no theoretical relation which exists over a wide
range.
(u) The difficulty of maintaining the cold junction at a fixed
temperature throughout and the difficulty of applying necessary
correction if it be not at 0°C.
{iii) Parasitic electro-motive forces developed in the circuit due
to Peltier and other effects.
(iy) Leakage from the mains or furnace circuit. The presence of
leakage current can be detected by short-circuiting the thermo couple
when the galvanometer continues to be deflected.
(y) They are not so accurate at low temperatures.
Applications- (*J They are widely used in industry for measuring
temperatures of gas, blast and other furnaces.
(uj They can also record temperatures of heated surface under
working conditions in paper, textile and rubber industries.
(ii} Total radiation Pyrometers. In these instruments the total
energy of radiations emitted by a source is measured a^d tne
temperature deduced by the application of Stefan's Law. Fery s total
aadiation pyrometer is an example.
THERMOMfimY .
I. , • . I . ^
{*) Fery's total radiation pyrometer. A standard form of
radiation pyrometer designed by Fery is shown in the Fig, I 2.1 3. It
consists of a concave mirror M made of copper with front side nickled
and having a small opening H in the centre. In front of the mirror is
placed a small diaphragm D backed by a blackened receiving surface.
One junction of a thermo-couple is attached to the back surface
and the other junction is shielded from the direct radiations by means
ol tonpe and a box. The leads from the thermo-couple are
joiiwd to a millivoltmeter mV and the electro-motive force developed
IS thus determiued,
The mirror M can be moved by a rack and pinion arrangement so
opening in the furnace on the diaphragm
IS correct, the two halves of the diaphragm are
^ mirror. These halves are inclined to each
of an diaphragm is viewed with the help
forte through the opening in the concave mirror. When
foJm a r® the images of the two halves overlap exact^ to
be d?dt^®S\S^ distance from the furnace. ^This residt cTn^easilv
the furnace ^be® doubted between the mirror and
rnace be dopbted. the total amount of heat radiations
?06
ENGINEERING PHYSICS
pceived by the concave mirror is reduced to J according to the
inverse square law. But at the same time the area which the image
covers is also reduced to one-fourth. Hence the amount of radiation
received and the area over which it is concentrated decreases in the
same ratio. Thus we see that the amount of radiations received per
unit area per second is constant and so the reading of the milli-
voltmeter remains unchanged.
It is a simple matter to find the relation between the size of the
object (furnace aperture) and the distance for this type of p 5 Tometer.
When the furnace is at a given distance the size of the furnace
aperature, necessary to fill the opening in diaphragm completely, can
be calculated from the geometry of the Fery's optical system and
hence is adjusted accordingly. After this adjusiment, the readings of
the instrument become independent of its distance from the object.
Calculation of Temperature.
Let Tq be the temperature of the hot junction 8.
T be the unknown temperature of the hot body.
V be the reading of the milli-voltmeter.
Then by applying Stefan’s Law, reading of the voltmeter is
given by
F=a(r6— To&)
where 6 is a constant and varies from 3.8 to 4.2. Because of this
variation from Stefan’s fourth power law, the instrument has to
calibrated by sighting it on a standard hot body which is raised t6
various temperatures. The temperatures are measured by a standard
thermo-couple and the corresponding readings of the millivolmater are
noted. The caliberation curve is thus drawn and the unknowh
temperature can then be read from this caliberation curve.
Advantages, {a) The instrument is direct reading and quick
acting.
(a) It is robust and has arragement to render the readings
independent of the distance between the instrument (pyrometer) and
the object (hot body) within certain limits.
(ii) Disappearing Filament optical Pyrometer. I" ,
pyrometer the intensity of radiation of a certain wave
by a hot body is compared with that of the radiation o the same
wavelength emitted by a standard body. The ^
until the two intensities are equal. The temperature of the hot y
must then be equal to that of the standard body.
The pyrometer as shown in Fig. 12.14 consists of “ telescope havmg
an obiective lens at one end and an eye piece _E at the other
end. An electric lamp having a straight filament F is placed m ^
THERMOMETRY
307
position of cross-wires with limiting diaphragms DD^ on either side of
it. The filamant of the lamp can be heated to various tampeatures
by adjusting the strength of the current passing through the filament.
The current in the battery circuit is measured with the help of a
sensitive milliammeter A.
HmATtOH
FRO^ THE
HOT body
RHEOSTAT
Fig. 12.14.
^ adjusting the position of the objective the
thr. the filamedt Tthe lamp
Pl"te of red glass iJ is placed at the p3„
the image of thl'^oulce'^iHhe aPetat brighter than
2s.;x'='“r ? =f‘ iStSTt-thrs. s;- :£
Kr4 “S^SaTe? Th". S“ lift
is,“ht;r“ s“r ■
The pyhe^g.„...
which can be determined by c^iberati^^^l?^* ■*' + ^ constants
r«~« I th.‘^‘e!Si?5.T.‘3'S.„T"S,i‘"'
ao9
EhJGINEERING PHYSICS
For industrial work, the mUl-amroeter may also be,caliberated
directly in degrees by focusing the instrument on a black body which,
can be maintiand at various temperatures measured by sensitive
thermo-couples.
The normal range of temperature which can be measured with
this instrument is 600°C to 1500®C as beyond the maximum limit
the filament rapidly deteriorates.
ModificatioD for measurement of very bight emperatures. For
the measurement of temperatures higher than
the normal range a rotating sector is
employed to cut out a known fraction of the
incident radiations. Suppose an opaque disc
from which a sector 6 is cut off as shown in
Fig. 12. 15 is rotated about the axis of pyrometer
tube between the hot body and the objective
of the telescope. If is the temperature
measured by the , instrument, then according
to Stefan's Law, we have
Fig. 12.15.
If is the actual temperature, then
Ti* OL 360
T*
$
• »
TJ ~ 360 I
• •
T.
Suppose 6=22-5® and the temperature
instrument is isOO^iC, then
Ti indicated by the
4
22-5
' =1500 (I6)i
= 1500 X2=3000®K
12.8. International Scale of temperature. We have ^een above m several
places that different types of thermometers ^ave different seal t
temperatures and no scale of To overcome
throughout its range except at tlie two fixed PO*"*®-
this disadvantage in accurate scientific should be
dered essential that some particular scale of ocurate
chosen as the standard. For achieving 7 'form.ty m the^ acc^um ^
statement of temperature for scientific "'ork» aHnoted in
by international committee of Weights and international
1927 a practical scale of temperatures known as the i
temptrainre scale. On this scale a senes of by the
pointsof substances in a given state of ^ points, which
constant volume hydrogen thermometer. specified
are conveniently and accurately of temperatures
merfns of interpolation, provide a practical scale F
rrHERMOMETRY - iJW J
309
which gives the closest possible realization of Thermodynamic scale
and at the same time permits of uniformity of temperature statement.
The temperature were designated on this scale by ""C" or
(int)". The introduction of this scale has facilitated the task of
standardising thermometers by the standard labora;tories all over the
world. :
The particulars of the instruments required and the formulae to
be used to realize the temperatures between the fixed points have
also been specified.
Basic fixed points^
(а) Boiling point of liquid oxygen
(б) Melting point of ice
(c) Boiling point of water
{d) Boiling point of Sulphur
(c) Melting point of Silver
(/) Melting point of Gold
— 182-970®C
O-OOO^C
ioo*oo°G
444-6^0
960-8^0
1063 0®G
The scale is divided into four parts for the purpose of inter*
polation. The methods of interpolation are briefly given as under.
(i) From 0®C to 660®C. Measurements are made by platinum
resistance thermometer by using the formula
{l+af+p(*)
^ The constants Rq, a and p are to be determined by measuring the
resistance of the thermometer at the ice, steam and sulphur points
respectively. ^ ^
(ti) From — 190®C to 0 ®C. The temperature is deduced from the
resistance Rt of a standard platinum resistance thermometer in ft)
above by means of the formula. ' ^
Rt — Ro [I (< — 100)
TesisI!‘nJ°nTfh"lh^°' ^ are determined by measurements of
points ' ^ thermometer at the ,ce, steam, sulphur and oxygen
^orih tVoth« being‘oraT;r/ot 9°0% pUtin^L" and
of eich wir‘e"murt Te'"b:;wel"'‘o'?3riKls" mm'^I^*"
given by the formula ® Temperature is
SIO
ENGINEERING PHYSICS
=2Sa?!"— ‘ p=
ki.!i U wave length emitted bv thp
fnlln ^ point. The temperature is determined by the
following formula known as Planck's radiation law. ^
log r^--
1
1
A L1336 (< + 273) J
where is a constant taken as *1-432 cm -degrees.
Expected Questions
are given to VheS Do^ntl°orr^°? generally used ? What numbera
-ic given ro tii^e points on centigrade scale and Fahrenheit seales ?
a thermometer^nf P^^^JJ^/^^'stance thermometer. Hew would you caliberate
?e tt“‘Lo“rter''^ra.e7““
Pvromater^''?^WhY®h ^ Total Radiation
measuremenf ? upper limit to its maximum temperature
measurement ? Suggest a method for extending its temperature range.
ature ?7an irbe'ek‘nde7? How What is its raage of ten.per-
5. (a) Explain the following. :
(i) International scale of temperature.
(ii) Choice of metals in thermo-electric thermometer-
(h) What types of instruments would you use for the accurate measurement
of temperatures in
(i) a boiler furnace
(ii) a steam pipe carrying highly super heated steam.
CHAPER XIII
THERMODYNAMICS
13.1. Thermodynamics. It deals with the behaviour of gases and
vapours under variations of temperature and pressure, and any
process that produces a change in the properties is known as
Thermodynamic 'process. For any process to be performed, a working
substance is essential, which is termed as thermodynamic medium. In
case of steam engines this working substance is water or steam,
where as in case of internal combustion engines it is air.
[Note. The \yorking substance is the agent which enables the
heat engine to do its work. In principle, it does not matter what
sort of fluid it is.]
\3;.Z-^irst Law of Thermodynamics. This law may be stated in two
slightly different forms.
(t) The first form of this law is one which establishes equivalence
between mechanical work and the heat energy. To be more specific
it can be stated as follows :
‘When transformation of heal energy takes place into the mechanical
energy (work), the amount of work performed is proportional to the amount
of heat energy which disappears, and if the transformation occurs in
the opposite direction, the heat energy produced is exactly equivalent to
the mechanical energy consumed.*'
For example, for producing one calorie, of heat 4-2 joules of work
always be done some how. In general if W is the amount of
work done and H is the corresponding amount of heat produced
then W^JH
where J is the mechanical equivalent of heat and has a value given by
«/=4*2 joules/cal.
J=4-2X 10 ’ ergs Mai.
=4200 joules/k. cal. _
J=778 ft. Ibs.'B H.U.
, T.- . .^=1400 ft. Ibs./C.H.ur^-
crea\eii nlVde:troyfdTuTX t ""
this general, law. it is stS^that!’'^ wwemrfwn of energy. Extending
X
3i2
ENGINEERING PHYSICS
"In all transformations, ike heat energy supplied to a su bstan ce is
equal to the increase in internal energy of the substance plus external
work done by the substance.
It can be symbolically stated as :
dQ=dE-\-dW
dQsaheat transferred
(Zj5=change in internal energy [m dT)
dR''=the external work done (JPdr)
l^T^^^econd Law of Thermodynamics. The first law only gives the
equivalence between the quantity of heat used and the mechanical
work done. But it does not give any idea of the extent to which the
transformation of one type of energy into another can proceed. This
is given by "Second Law of Thermodynamics" which has been enunci-
ated by Clausius and Kelvin in slightly different words.
(а) (i) Clausius’ statement. "It is impossible for a self-acting
machine working in a cyclical process, unaided by any external agency to
convey heat from a body at a lower temperature to a body at a higher
temperature."
{«*) Heat cannot flow from a cold body to a hot body without the
performance of work by some external agency.
(б) Kelvin's statement. It is impossible to derive continuous supply
of work by cooling a body to a temperature lower than that of the coldest of
its surroundings.
The above statements give that no actual or ideal heat engine
operating in cycles can convert into work all the heat energy supplied
to the working substance. But whatever the conversion niay be
there, it will essentially be in the ratio of the mechanical equivalent
of heat from the first law. This leads us to define the thermal
efficiency.
Thermal Efficiency. It is the ratio of the heat converted into
useful work to the heat supplied.
Thus thermal efficiency
Heat converted into useful wo rk
Heat supplied
Heat supplied — He at rejected
Heat supplied
Heat rejected
Heat supplied
Most efficient heat engines i.e., *+ rhmp etc has
have thermal efficiencies much above 40 A. Steam
an efficiency of nearly 30 to 35% and steam engine 8 to 10 /o-
ITHIAW'ODYNAMICS . 313
• «
13 ’ Reversible proce^M and irreversible process.
» ► ' » *
-Re^rsible process. A reversible process is that which can be
■retraced in the opposite direction so that the working substance passes
through exactly the same conditions as it does in the direct process.
(b) A process in which this does not take place is called an irre-
Tersible process.
Examples, (t) Reversible process, i. All isothermal and adia-
batic changes are reversible when perforpaed slowfy. In such a case
it is assumed that there is no friction to be overcome as work done
in overeoming friction is wasted. The pi'ocess jcarried out is very
slow, hence no energy is wasted in producing oscillations and eddy
currents and hence no heat is lost by conduction, convection and
radiation.
Under these conditions if heat is supplied to a given mass of a
gas at constant pressure, it expands and does some external work.
It the same amount of work is done on the gas in compressing it.
the same quantity of heat will be given out.
2. Ice melts when a certain amount of heat is absorbed by it.
ft water so formed can be converted into ice if the same amount
of heat IS removed from it.
be nohJl ^ thermocouple is neglected, there will
effect. In such a case
effect '2''ersible. At a junction where a cooling
an equal heatl ® ff/f -‘" f direction
an equal heating effect is produced when the current is reversed
wire is irreversibkTUs^because^even^whA ^ resistance
same effect is observed. ^ current is reversed the
mustbesatlsfiS'^ll^^^”^®**^ Possible the following conditions
sta£be?„Vhtmodrn^fc%"qr^^^ at all
condndtionrconvectfon ®a^S “lu?ronlc"°
314
ENGINEERING PHYSICS
working substance is made to pass. If these operations are arranged
in such an order that the substance is regularly brought back to its
initial state, then these operations constitute a cycle. When the
series is complete, the cycle may be repeated any number of times.
If the operations of a cycle are plotted on a P—V diagram, they
form a closed figure, each operation being represented by its own
Fig. 13.1.
curve The net amount of work done by the working substance
during one complete cycle is given by the enclosed area of the P~V
diagram shown shaded in Fig. 13.1.
The ideal cycle is one in which all the operations are reversible.
1^6 The Heat engine. The heat engine is a mechanical contrivance
whichenablesafluid. called the working substance, to go through a
CTcle of operations, as a result of which the mechanism can perform
useful external work. As per second law of thermodynamics only a
uoftion of the heat supplied to the engine is converted into useful
^ t the rest is rejected unused. The temperature at which heat
^nnhed to the working substance is always higher than that at
IS suppl ed to the worK ng or aeurce ^eat to
The heat rnginfu called the Heat source, and the cold body >nto which
the heat engine IS ca thi It is not possible dt
ratures, at least no one temperature
conversion of heat into work y^ working substance
(used in tiding working substance as
rr;Sn^Tlnter"nllto^^^^^^^^ -d ^in motor cars.
aeroplanes etc. , „ .
Thus there are three essential parts of a heat engine. ( )
THERMODYNAMICS
315-
.ource (») Working substance (Hi) Sink. In case of steam engines
etc. these are the Fumace, stcafn and the condenser, and for inteinal
combustion engines heat is produced by a mur^ttrc of air and
vapour, the working substance as already stated is oir and the
sink IS the atmosphere.
engine is, of course, a very complicated machine
clr “^o'sl?" • ^ <=o"'P«<=t unit such as the engine of a motor
rrinv ^^°^®"g‘nes which require two complete revolutions of the
co^letP «ug‘"es* where as some require one
T-stioke ensinrs°" completion are known as
th=.+T^® regarded as a frictionless machine so
a?Ue friction does 'not
of TeaMnlo wor^t^H are studied, but only the transformation
oi neat into work and work into heat. A French engineer SaHf
Carnot in the early part of the 19 th century proved that even with an
\ff engine was not possible to convert nJe Ln a certaZ peTentZ
was able to establish these pSn^r ‘
^Vim?er;!o'n^''oVactu^a?r>^ f--
and hence never realised in practice,
was considered by Carnot for theoreti-
cal investigation as follows. The
Ideal engine consists of ;
“In'i*”^ ioflom. In it mot?r a
perfectly non-conducting piston in a
horizontal direction without friction.
The cyhnder contains air {which is
supposed to behave like a peXf ga
working substance and is
placed on a perfectly insulated stand
(**) A. hot body to serve as n
= 5(.tlofu\e.^ ~
®®rve as sink
aL^olute!'^°^ temperature
Fig. 13.2
temp^a*m““ ma1n*“ra«rcaJfc®!m'^®^^^^ so that thei
heat to or from the cylinder. ^ changed during any transfer
ENGINEERIMG physics
Let the cylinder contain m gm. of air and let its original condition
be icpresented by point (I) on the P — V diagram as shown in Fig.
13.2. In a Carnot’s cycle the working substance is supposed ’to
undergo the following four operations : —
[a) Operation 1 — 2. Let the initial temperature of the gas
the Cylinder be Ti°A. Place it in contact with the source and let the
piston move outward very slowly. As the piston moves, the gas
expands and the temperature tends, to fall. Heat is, therefore
absorbed from the source at a constant temperature Tf The cha^e
is isothermal and is represented by the curve 1 — 2 in Fig. 13.2. The
.amount of heat absorbed Qj from the source is equal to the work
done Wx by the gas in free expansion from a pressure pi and volume
^Vi at (I) to pressure volume Vg (2)*
/ ... lFj=Work done by the gas or air
loge ^ (already proved)
F,
mRTx loge ^
0:
or
Qx^mARTx ^oge ^
(b) O Deration 2— 3. Now remove the cylinder from the source
and pllce^Ln the perfectly insulated stand and al^ow ^he ga^o
expand adiabatically till its temperature falls ^ ^ fs
is represented by adiabatic (2-3). The work done W, by the gas
given by the relation
«7 (as already proved)
1 y— I (y — 1)
lr\ Oneration 3—4- In order to bring the working g=»s
its original'pressure and volume remove the cylinder from he -
lated s^tand^and place it on the sink at ^ t^Tomp^eted anl
the piston very slowly inwards so t^at the g ^
the temperature tends to rise. Heat IS, therefore ^ repres-
constant temperature TaM. T ec ange IS ^
WQ=—mRTz loge
= — mRTz loge
—mART^ logtf
V*
Vg
n
TOERMODYNAMICS .
317 '
1. Remove the cylinder from the sink and again
place It on the insulated stand. Allow the piston to move inwards
so that the gas is compressed adiabatically until the temperature rises-
^om to and the gas attains its original pressure and volume.
The change is represented by the adiabatic 4—1.
Work done on the gas during the compression is
y -1
— PiVi —PaP*.
y-1
or
or
(y-i)
-
(y-l)
13.8. Efficiency of Carnot’s Cycle. It will be noticed
done by gas during adiabatic expansion (2— 3 ) is equal
done on the gas during the adiabatic compression (4—1)
that the work
to the work
♦.e.
Hence net work done=Pri 4 -Fr^-ira-Tr 4 =fri-ir,
Heat in put to gas is Q^^mART^ log^
^1
Heat rejected by gas is Q2^mART^\og^ %
The net amount of heat absorbed in units of work^Qj— .Q
Net work done by gas is equivalent oi Q^~~Q^ *’
Hence Efficiency=
In put
or ^^ .^ount of heat co nverted into
Total amount of heaFabsorbed '
Since the points 2 and 3 lie on the same adiabatic
Similarly points 1 and 4 lie on the same adiabatic
The points 1 and 2 lie on the same isothermal at
PxVl-^PtV^
Similarly the points 3 and 4 lie'^^he same isothermal at r
Pa^s^PaVa ^
ENGINSRtNG PHYSICS
Multiplying corresponding sides of the above expression we get
V V
—v/ v^v/ V4
or
— 1
or
^2 .
=i!3
^X
V 4
Let
h.
D
II
V4
{r
is isothermal
•
Qx
—mARTx loge
V;
^4
Q^=mART^ loge ——'mART^ loge r
Now
Qi—Qi^rnAR loge r
^^ Qx-Qt ^mAR (Ti-jyjo^
Qx
T.
mARTi loge r
or - -• ^=1-^
which shows (i) iAe efficiency depends only upon ike temperaiure of
Ike kot and cold bodies and is independent of tke nature of tke woTKxng
substance.
If for a Carnot's engine
jri=273 + 100=373°K
and 2^2=273 + 0 = 273®^
^_ ri— y2 _ 373 — 273^ ^10j
then,
lOO
373
373
o/oage , = 100 = 27%.
It shows that even in an ideal heat engine operating on an ide^
cycle the maximum efficiency obtanable
fraction of the heat energy has been conver ed accordance
73 <y of the heat energy remains unavailable, whicn is
with Second Law of Thermodynamics.
iii) 100% efficiency can be achieved only when ^reac es
absolute zero although it is impossible to achieve P
13.8. Internal Combustion Engines. ^"®‘"^|n^^*^whic^onverts
an internal combustion engine is also a heat g
thermodynamics
di»
part of the heat energy of fuel into mechanical work. In this engine
the combustion of the fuel takes place internally that is inside the
engine cylinder itself. - Consequently, very high temperatures are
produced in the cylinder which, if not kept down to proper values
will damage the metal of the cylinder and the valves. As the high
temperatures are produced, the thermal efficiency of an internal
combustion engine is very muc i improved to 35 to 40%. Also
because of the absence of auxiliary units like a boiler and condenser
etc. which are in a steam engine, the unit is a compact one. It is
because of these considerations that internal combustion engines have
become so popular in all spheres and are in extensive use in numerous
fields like auto-mobiles, locomotives, generating sets, submarines,
ships, aeroplanes, and in several industrial units. Internal combustion
engine has largely replaced the steam engine and at present time the
steam turbine alone, is comparable to the internal combustion engine
in the amount of horse power produced.
As mentioned in the above lines that in case of internal combus-
tion engines the fuel is burnt inside the engine cylinder and it is made
to give up its heat to the working substance i.e. air; The fuel used
may be petrol, gas or oil. These engines can be classified on any
of the following bases : —
(а) Spark igixition (S.I.) Engine.
(б) Compression ignition (C.I.) Engine.
♦ •
(а) Spark ignition engines- In these engines petrol, paraffin and
gas are used as the fuel. The charge admitted to the engine cylinder
is a homogeneous mixture of petrol and air which is delivered in
correct proportions depending upon the load on the engine'''by a
popular device known as carburettor. The mixture is compressed in
the cylinder and is ignited by an electric spark from the spark plug.
These engines work on what is called an Otto Cycle.
(б) Compression ignition engines- These engines use heavy oil
which can not be easily vaporised. The oil is introduced into the
cylinder in a very fine spray, mixed with air. It should be noted
that the ignition is not being effected here by an external agency as
in case ot Olio Cycle. It is entirely due to the high temperatures
produced by the high compression of the air. No electric spark is
needed for ignition. These engines work on what is called a Diesel
Cycle.
l^SK'^amot’s Theorem.
Assuming the truth of the second law of thermodynamics it can
be shown that no heat engine is more efficient than a reversible engine
when tvorhng between the same two limits of temperatures. This is known
as Carnot s theorem.
Suppose we have two heat engines A and B working between
the same two limits of temperatures T, and r,, i.e., heat is^ absorbed
S20
engineering PHYSlCSf
at a temperatute from the source and is rejected to the sink at
a temperature 0. Suppose A is more efficient than B, then ia B
we may suppose that the length of the stroke is so adjusted that
the work performed in a complete cycle is equal to that performed
by A under similar circumstances.
Let A be coupled with B, by a suitable machinery so that
token it works directly, it drives B in the reverse direction as shown in
Fig 13 3 Then during each complete cycle A will
absorb a quantity of heat at a temperature
and reject a quantity at a temperature
t\ On the other hand when B traverses a
cycle in the reverse direction, it will absorb a
quantity of heat Q/ at a temperature and
will reject a quantity Q/ at a temperature
T Thus the amount of heat energy converted
into work by the engine A, in the first case
Q%
and amount o{ heat energy converted into work
by the engine B, in the second case
=ei' -Q,'
Since the worksjperiormed in both the cases are equal
Qi,—Q,=Qi Qi-
Now by supposition the efficiency of A is greater than that of B
or
or
Similarly
Qi--Qt^Qi-Qz
Qi
l.>-^
Qi Qi
Qi>Qi
Qi>Q2-
Similarly . .
This shows that the quantity of heat Qi i absorbs
the source is greater than the quantity of which
tnTce Tt a higher temperature T,. Smce no external
Supposed to aid during the whole process P\"“*temperature T,
temperature r, to^a^body ^a^t a^^h.g^^ „
.h. ... oi ^
rey^sible engine.
♦
Now the efi&ciency of a reversible heat engine is ^ l — ^ ^
when working between the temperatures T, and T^. This is inde-
pendent of the nature of the working substance. Hence we see that
tU efficiency of all reversible heat engines working between the same
lifMts of temperatures is the same, which is Carnot's theorem.
aausins-Clapeyron's Eqnation. The boiling point as well
point of a liquid depends upon the pressure. It is
possible to obtain a formula showing how the boiling point and
the application Sf the second
law of ther^dirnamtcs. The relation thus obtained is known as
Llausius Clapeyroo s equatioo*
as thr^wornL^^ir." ® ""'t ® liquid
Tnn l substance in the cylinder. Let aABh and
the two isothennals for
this substance at temperatures T®
andT‘>^dr« Absolute. Up to A
the substance is wholly in a liquid
state and along AB it exists partly
^ i Partly ih a vaporous
state. The temperature T is, there-
fore, the boiling point of the liquid
at a pressure corresponding to A.
Similarly along the isothermal CD
the liquid wUl be partly in a liquid
and partly in a vaporous state at
a temperature Absolute.
1 he temperature T^dT represents Fig. 3.4.
the boilmg point of the iiquid at a pressure corresponding to D
Let the volume of the liquid at A be c.c. Now
liquid slowly gainVhiLt'^and h*Mnv”ted temperature T. The
constant pressure. The process is^Iothe^ vaporous state at a
AB. At B the whole of the liquid is Jonverted^"^**® represented by
Krnt''o 1 rat'^^^^^^^^^^ “eat ^Tth^n"
“veb^®" 11*° e^e"d adiIba"ticl*Uy srthat“Sfm®
indicated by the point Z) is reached'. The Proc^s^'^isotherr^^^t a
rCCVM£ ^
322
ENGINEERING PHYSICS
quantity of heat L^dL is given out at a temperature T — dT, Here
Ij — dL is the latent heat at this temperature.
(iv) Place the cylinder on a non-conducting stand and compress
the liquid so that the initial volume and temperature as at A are
attained. The process is adiabatic and is represented by the curve
DA.
This is a reversible cycle because it can be performed in the opposite
direction and each change in it can be reversed.
The efficiency of a reversible cycle is given by
Qi — Q2 ^2
Qi T,
L~{L-AL) _ T- {T^dT)
L T
L T
The quantity of heat converted into work=Zf— (£ — dL)=dL
Work done in one cyc]e=area ABCD
s=dp{v ^ — work units
in heat units.
«/
If dL is measured in units of heat, then
J
Substituting the value of dL in equation (i). we have
dpK— Vi)
JL
...(n)
T
or
dp JL
dT ^(^2-^2)
(1) Effect of pressure on boiling point.
U.. there is an increase in volume when the liquid “^nges
into its vaporous state, as is always the case, then {v^ i) P
Since L is also positive.
^ is positive.
This shows that the boiling point of a liquid is ra%aed by increase f
pressure. ,
( 2 ) Eflfect of pressure on the melting orthere may
there may be an increase m volume as m the case ot wax
,be a decrease in volume as in the case of ice.
THBIMODYNAMICS
323
In the case of melting wax (^2 — *^i) positive, therefore, the
tnelting point in this case is raised by increase of pressure as ts
positive.
In the case of melting of ice is negative, hence the melting
point is lowered by increase of pressure.
In general the melting point of those substances which expand on
■melting is raised by increase of pressure and vice versa*
13.11- Otto Cycle
The present day petrol (gasoline) engine operates on this Otto
cycle because it was introduced in proctical from by a German
scientist Otto. This is also known as constant volume cycle.
The P-V diagram for a theoratical Otto cycle is shown in the
Fig. 13 5.
The cycle consists of two constant volume lines and two
adiabatic curves as shown. Let at a point 1, the cylinder be full
of air of mass m gm. which has a volume of Vi pressure of
and absolute temperatur of The cycle is complated in four
stages described below.
(i) First Stage. The bottom of the cylinder is covered with the
insulating cap and the piston is moved inwards compressing the air
adiabatically till it reaches the end of its stroke at point 2. The
pressure and temperature of the air rise to P, and respectively
whereas its volume is reduced to a value of Work is done on th&
air during this adiabatic compression.
tig. 13.5.
324 ENGINEERING PHYSlC$
rl ‘.l'- ‘ • •' 1 ' >
(it) Second Stage
At point 2 , insulating cap is removed, the cylinder bottom is-
brought in contact with the hot surface and then piston is held
stationary. The air is heated at constant volume till its pressure and
temperature rise to the values of Pq and respectevely. The heot
taken in is
-Hi=wOt,(T3 — P 2 )
{Hi) Third Stage
At point 3 , hot body is removed and the insulating cap is again
put on the bottom oi the cylinder. The air is allowed to expand
adiabatically till point 4 is reached. During this expansion, work is
done by the air whose pressure and temperature fall to the values of
P 4 and Ti. respectively.
{iv) Fourth Stage
At point 4 , insulating cap is removed and the cylinder bottom is
brought in contact with the cold sink. The pistion is held stationary
and the air is allowed to reject heat to the sink at constant volume
till its pressure and timperature drop to their original values of Pi and
T, respectirely. The omount of heat rejected to the sink is
The cycle is thus completed because the air has been brought back
to the same conditions of pressure, volume and temperture as at tne
beginning of the cycle.
Ideal Efficiency
Heat received (^ 3 — '^z)
Heat rejected H 2 =mCv — ^i)
Work done =Hi—
I
Efficieney
2
T^-Ty
Now from adiabatic compression AB, we have
tHERMOtSYNAMlCS
325
Similarly from adiabalie expansion CD, we have.
/n * T7 .. Y 1 :
(2) we get :
Since ^
...( 2 )
> hence from'Equations (i) and
T^~ T, - r.
or
T. * r, ‘
or
f
1
1
or
T, T^-T,
Tt ^T^-Tt
Now
7-1
V ■ - '
Since -f^s=adiabatlc compression ratio p
Hence
s-(^)
y— 1
...(3)
Substituting the value from Eq. ( 3 ) we have-
Air-standard Efficiency 7=1
Air«standard
y-1
—(f)
diagram o'f shown K’ theoretical P-7
modified when used in an internal comH '*'*'^*
petrol or gas. The cyc?e in^Wes sS nro°“
four require movement of the oiston “"t of which
The engine is known as four-stroke enSlT strokes,
engme consists of a pistion Ld a cvUnrt~® •*" W
inlet vailves and exhaust valves wl^ich is provided )vith
usivaives. The SIX processes are as follows :-:-
ENGINEEKIKG PHYSICS
S26
INLET EXHAUST
(i)
inlet exhaust
(ii)
•<LET EXHAUST
(iii)
INLET EXHAUST
(iv)
Ifig. 13.6
(i) Suction or Charging Stroke plosive
E»r£. “
centre (BDC). This is shown by line 1,2 in tne
(ii) Compression Stroke ^jvtureis
In this stroke, all valves are closed and the ®^^^g^''pj.essure and
.™p,.s«d f l.b‘f ■"J’ ™ S'n.id.P.bly. Tl,. pUon
SchS“?p””«i i«." TO. p.«~ 1. •“
thermodynamics
327
Explosion At or near the end of the compression stroke, the com-
bustible mixture is fired by a series of electric sparks from outside.
It burns with explosive voilence thereby releasing heat energy. There
Fig. 13.7.
is a rapid rise in the pressure and temperature of the products of
combustion but their volume remains practically constant. The
piston does not move in this process. This combustion is shown by
the curve <70 which is practically a straight line.
(m) Working or Power stroke
Due to the explsioon of the mixture, the piston is pushed out
with great force. The hot burnt gases expand, the expansion being
approximately adiabatic due to the high speed of the piston. During
the expansion DE some of the heat energy produced is transformed
into work. The pressure and temperature of the burnt gases fall to the
values represented by point E. Even though there is a good deal of
energy in the hot gases, yet no further energy tronsformation can
occur since the piston cannot go further (because it has already
reached its BDC) ^
Exhaust, At the end of the power stroke i.e., at point E. The
exhaust valve opens and allows gases to escape until the pressure
drops tothat of the atmosphere. Heat is rejected to the surround-
mgs at practically constant volume (line EF)
(tv) Exhaust Stroke
which the piston, while
moving in wards, pushes the remaining burnt gases out The cycle is
an? « the next cWg^of ak
smd ^trol vapours or a mixture of air and gas. It is shlwn by
Eximple l. A gas ^ne vioriing on the otto eycU has a cylinder
0 } ieameter 152-4 mm and a stroke of 22l:-6 mm. f/the Oeardnee whS
» •
328
.(
. . I
ENGINEERING PHYSICS
is 1310 cm^, find the air-standard efficiency of the engine,
Swept volume
TT
= — Z)2 X stroke length
_ 7cx (15-24)^
4
\
Total volume
X22*86^t6(| cm®.
= Swept volume +c].ekrahc^ vblume
=4I60+1310=S4?0 cn>^ 1
Compression ratio, p = ?^*^=; 4^'8
/
^ K
<.
Air-standard
( > 1 vr— J
ndard -^=-1 ^'^7 " —
— (^Vs)'
1*4—1
=0*43 or 43%.
Example 2. The efficiency of an otto cycle is 50% and y is 1'5.
What is the compression ratio ?
Now
’7=1 —
kt)
y-1
Substituting the values
50
=(
1
r^)x
100
100 «0*5
or
0*5=1
1
0*5
or
1
0*5
=0*5 or P=(2)
0‘5
\ '
= ( 2 )
=4
r
Compression ratio=4I
« k
V , ^ fck'. .
\ . • t
c;
'tHHlMODYrtAHlCS ’
329
Examples. Derive an expression for a change in e^ciency for a
change in compression ratio. If the compression ratio is increased from
S to 8, what will be the percentage increase in efficiency ?
Air standard efficiency
where
Let
y=. 1 *4
pj=original compression ratio
f> 2 =final compression ratio
1
Original efficiency — I
<>/
1
Final Efficiency
y-1
Percentage increase in efficiency i
P* ' •
1
1
y-1
Now
I
Pi=6
P.b8
c
2
_r
-)
j
Percentage increase in efficiency
'I
X 100
— 1
r,-
XlOO
1
8
0*4
1 1
I
= 8 %.
1
0*4
1
I
— 1 1 XIOO
I
J
• C« • i .
ENGINEERING PHYSICS
SdO
13 . 12 . Theoretical Diesel Cycle
- , Diesel Cycle, also known as the constant Pressure Cycle was
first introduced by Dr. Rudolph Diesel. It consists of two adiabatics
one constant pressure line and one constant volume line. The engines
used in heavy motor vehicles, stationery porwer plants, big industrial
units and marine ships etc. use this cycle of operation.
The different stages of the cycle are shown on the PV diagram,
lo begin with let us suppose that the engine cylinder is full of air of
mass *771 gms which has a pressure of Pi, volume of F, and temperat-
ure of as represented by point l.
First Stage
The piston moves inwards compressing the air adiabatically
(according to the law PF'-^constant) upto point 2. The air now
occupies the clearance space. The pressure, volume and temperature
rise to Pg, Fg and Pg respectively.
Fig. 13.8.
Second Stage
At point 2, heat is supplied to the compressed air from an
external source of heat at constant pressure till point 3 is reached.
The amount of heat absorbed is Qi=m Cp {T^ — Ti)
Third Stage
At point 3, the heat supply is cut off and this point is known
as the point of cut off. The The air is allowed to expand adiabatically
(according to the law PF'’=constant) till point 4 is reached.
Fourth Stage
At point 4, the hot air rejects heat to the sink at constant
volume. The piston is held stationary during this process
pressure falls to Pj. This completes the cycle, since air is restored to
its original conditions.
TEHRMODYNAMICS
The amount of heat rejected is given by
Q^=mOv {Ti — Ti)‘
13.13. Air-Standard Efficiency of Diesel Cycle
Heat received Gp (^s ^a)
Heat rejected out Q 2 =^Cv (^* ^i)
Heat converted into work —Qi — Qt
Qx—Qi
Air standard efficiency =—g —
' Qi
mCx,
mCp {T 2 -T 2 )
_x 1
1 (TWi)
7(^3-'^)
Now Let ^ =adiabatic compression ratio=ri
and
and
Then,
Also
Also
^=adiabatic expansion ratio=r 2
^3
V2 ~ Tz
/ V ^
I',=2’i(y') =2', .{*•,)
Y-1
T. ^3 y
..[Adiabatic compression!
= 1» orT3«T,.^«=r,.a
T.~V
m
.[constant Pressure Process],
=a T,(ri)
Y-1
y-1
3’.=2’3(^’)
(A diabatie Expansion).
Y — 1
332
ENGINEERING PH YSldt
““■T'lW’' ' =7'.a^
Substituting the values of T^. and T^, we have
-r.
Ti.o r/ ' -T, r/
-'-ff
1
Thus air-standard efficiency of a diesel is
1
Engine The reciprocating steam engine is one of
Ihe oldest types of heat engine which converts heat energy into
mechanical work. It is an external combustion engine in which steam
■expands inside the engine cylinder doing work, on the piston.
Although the steam engine is being largely replaced by the modern
development of steam turbines and I. C. engines, it has still many
specialised applications. A brief description of a simple steam engine
is given below.
Fig. 13.9.
Reciprocatng Steam Engine
The fig. 13.9. above shows outlines of a double acting reciprocat-
ing steam engine. There is a piston which reciprocates in the cylinder.
There are two ports for admitting steam on either side of the piston
^nside the cylinder.Attached to the cylinder is the steam chest in which
She D-side valve reciprocates and alternatively opens and closes the two
THaMOI>yNAMICSr
333-
ports. The motion of this yalve is taken from the crankshaft by means
of aii eccentric,, eccentric rod and valve rod. Steam from the boiler
plant is led to the steam chest. Steam enters into the cylinder and
expands thus pushing the piston outwards. This thrust is transmitted
to the crankshaft though the cross-head and connecting rod. When one
stroke of piston is completed, the steam enters from the other side and
pushes the piston inwards. Like this the piston reciprocates the
reciprocating motion is converted into the rotory motion of the
crankshaft.
1315> Rankine Cycle
This is the standard cycle of operation which is employed for
estimating and comparing the performance of various external com-
bustion power plants. It is the ideal cycle on which a steam engine
works. The P—V diagram is shown below ;
Fig. 13.10.
Consider l kg. of water at an initial temperature of T.® absolute
^nd under a pressure of P, kg/cm« absolute. This state of water is
represented by the point A,
wa+ Let a hot body be brought in contact with this
rat/ water to an absolute tempe-
rature and a higher pressure P^.
is evaporated at constant [pressure P..
^ volume takes place. Latent heat is absorbed by water
ana work is done due to increase in volume.
exuan^^-TSI-*^ body is removed and steam is allowed to
P. and tA The pressure of steam falls to
e«on oKT.
brough in contact with the
Latent^KlA steam to the cold body.
t IS rejected out by steam which condenses back ' into
334
ENG INnRING PHYSICS
water at constant pressure. Eventually, whole of steam gets
converted into steam. Hence the cycle is completed.
The net work done by steam in one cycle is given by the enclosed
area ABCD.
t 3 . 16 . Efficiency of Rankine Cycle
Referring to the fig. J3.10. it can be seen that the heat is absorbed
during the operations *AB‘ and •BC* while it is rejected during the
operation *DA\
Heat Input^Heat absorbed during ^B-|-heat absorbed during BO.
Where Hwi and Hw^ are the enthalpies of water at B and A
respectively, the latent heat of steam while qx is the dryness
fraction of the steam
Heat rejected
Work done
• «
Efficiency
= Hwx + q^Lx Swx^Hx—
=sHeat rejected during DA
=Heat absorbed — Heat rejected
. * •
^Hx-(Bwx+qtLx)=Bx-Bx.
Work done _ Bx~B^
~*Heat Input Hx—Bv>^
^ H — J?
Thus Rankine Efficiency= ^
Note :-In actual practice, the cycle discussed above takes
place in the steam engine plant as a whole. The water is p p
from the condenser at a pressure />. and
boier at Ithe pressure Pj and is heated to steam at
(operation AB). Thia water is then raised by t*’® f elmThus
pressure P. and dryness fraction (operation T and
raised is then* expanded adiabatically to the temperature T. a^d
pressure P, (operation CD) in the engine cylinder In the end^me
steam is exhausted and condensed in the con ( P
thus completing the cycle. . ^ ,» ^nh/ntf
Example 4. tketkernu.1 f
hot and cold bodies have temperatwres of 2800 F (A.MJ-B. 1958)
The thermal efficiency of a carnot engine
Here y ,=2800 + 460 = 3260®iC
T^=^0+460—520°K.
PHBRMODYNAMIC
335
Substituting the values of and in the above relation we get,
3260—520
’7= ^260“ =®*8407 or 84*07%
Example 5. A reversible engine converts one-sixth of the heat input
into work. When the temperature of the sink is reduced by62^C, its
efficiency is doubled. Find the temperature of the source and the sink 7
{AM.I.E. 1959)
Since the engine converts Jth of the heat input into work
%.e.
T,
...( 1 )
Where is the temperature of the source and T. is the temper-
ature of the sink.
When the temperature of the sink is reduced by 62®C the
efficiency is doubled i.e. it becomes
T^-{T^- 62 )
- •••(«)
From (<), I—
£l_± *
Ti “ 6 ...(«»)
or
from fu), 1
J’a— 62
2’i
=\
or
from (Hi), T,= J-Fi
6
Substituting this in (it;), we have
=1
Solving the above equations we have,
2*1=372“^ or 90“C'
*'• ^1=310*^ or 37°0
the n-. land
iv for air is l-f) Calculate the air standard efficiency.
«
Let the clearance volume m*
336,
ENGlN^aiN9'PHYSlCS
then total cylinder volume ^1=13 w® .r ; j
and swept volume (Fi — Fj)s= 5 l 3 — 1 = 12
8
8 % of stroke volume {F 3 — F 2 )= j-^Xl2=0*96.
Hence total volume to point of cuf off=F 3
= clearance volume-|- 8 % of stroke
= l+0*96 = r96
Air standard efficiency =1
Y
or
17=1
-C-r)
— 1
(g^ - i)
y(a— 1)
y,
Vi
Vt
1-96
1
= 1*96
^1=tT- =
13
1
1*4—1
1*4
^ 1 ' - r(l’9 6* ^ — 1 ) ~|
v—^ ( 13 ) L l*4tr96 — 1 ) J
= 0-583 or 58-3%
»
Expected Qaestioos
1 , (a) Enunciate the first law of Thermodynamics.
(b) State and explain the significance of the second law of thermodynamics.
Give a brief summary of important implications of the second law.
2. What is Carnot's theorem ? How do you prove this theorem ?
3 Derive the clausius-clapeyron equation and show *^ 0 ^ with hy^^^
it. one <^n determine the variation of boiling point and melting point with p
variations. ^ , . j
4 . (a) What is Otto Cycle ? Why it is known as four stroke engine ^
lb) Derive an expression for its efficiency.
5 (a) Explain the four strokes of Diesel engine. Detive the expression
Its Compare the constant volume and constant pressure cycles.
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